Question 6: Simple Harmonic Motion

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Page Commencement date

Questions covered

Introduction

Formulae

Manipulation of equations

Questions involving ε

Introduction to Problem-Solving questions

Vertical Motion Exam Questions

Horizontal Motion Exam Questions

Guide to answering Exam Questions

************* Marking Schemes / Solutions to be photocopied separately ***********

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With the exception of the ‘Manipulation of equations’ section, all other questions will involve the following series of steps and so they need to be learned off.

I am just placing this here to make it easy to find the information when you need it – you don’t need to learn this now.

The extension of a spring or a string is the distance between the object’s Current Position and the end of the natural length of the string (in other words, how much the string is extended beyond its natural length).

The extension depends on the position of the mass, which changes throughout the question.OrThe extension continually changes as the mass moves over and back.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.). The amplitude therefore does not change.

N clucking B:The release point has no bearing on the question until after you have established step 3 (i.e. shown SHM). This is probably the most common mistake that students make when answering these questions.

Procedure for solving long questions1. Find Equilibrium Position (E.P.)

If we’re working in the horizontal direction with a single string then equilibrium position is at the end of the string.

If we’re working in the vertical direction then we use the fact that at E.P., Fdown = Fup

I tend to use the letter ‘d’ to represent this extension so that it doesn’t get confused with ‘x’ which we will use in the next section.

2. Show SHM

If we’re working in the horizontal direction (assuming just one string and no friction) then there is no small force. The big force will be the tension in the string so we can use F{ext) to represent it.

So for a string in the horizontal direction our equation will be 0 – k {ext} = ma.

If we’re working in the vertical direction then the small force will be mgSo for a spring in the vertical direction our equation will be mg – k {ext} = ma.

Before answering long questions check that you understand this by testing yourself on the last page.Introduction

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For a string in the horizontal direction, E.P. is at the end of the string

For a spring in the vertical direction the equation to find E.P. is: mg = k {ext}

To show SHM, extend the object a distance x beyond E.P. and use Fsmall – Fbig = ma

To show SHM in the horizontal direction use: – k{ext} = ma

To show SHM in the vertical direction use: mg – k{ext} = ma

Basically what’s going on is that an object which exhibits simple harmonic motion has its natural resting point (called ‘equilibrium position’) and when it gets disturbed from this point it heads back but overshoots the mark and has to return again and again.

In our study of Newton’s Laws of Motion we consider the motion of a body when subject to a constant force or forces. As a result we can calculate the object’s velocity or position at any time t.

However there are many instances when a moving object is subject to a changing force – can we still calculate future position and velocity? Well, we can if we can quantify the force, how the force is changing. One example of an object experiencing a changing force is a stretched spring. We say that the spring is undergoing Simple Harmonic Motion (SHM).

An object is said to be moving with Simple Harmonic Motion if;1. its acceleration is directly proportional to its distance from a fixed point in its path, and2. its acceleration is directed towards that point.

We represent this mathematically as:

The ‘fixed point’ referred to in the definition is also called the ‘equilibrium position’. The key to SHM is that the acceleration and the displacement are always opposite in direction. This is represented by the minus sign in the equation, although mathematically the minus sign has no

significance and can be ignored for calculations. We will ignore it from now on.

Other examples of SHM, apart from a stretched string are: A prong on a vibrating tuning fork A pendulum oscillating at a small angle The tides

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a = -2 x

What happens acceleration in Simple Harmonic Motion?Recap With vuast, acceleration is constant in magnitude (for a given stage). In SHM, acceleration by its nature is not constant in magnitude.

Let’s take another look at this.

Consider a Simple Pendulum

displacement velocity accelerationEquiplibriumposition

0 max 0

Extremeposition

max 0 max

The aspect which most people find confusing here is that at the extremities the velocity is zero, while the acceleration is a maximum, whereas at the equilibrium position the opposite is the case. How can this be?

AnswerYou must think in terms of forcesPersonally, I can’t fathom why textbooks don’t emphasise this more. After all, the only way you can have acceleration is if you have a force causing it. Acceleration is directly proportional to the force causing it, so if force is a maximum acceleration must be also, and vice versa. Now let’s look at the diagram above in terms of forces. At the equilibrium point the restoring force on the mass is zero – it’s not being pulled to the left or the right, therefore the acceleration must be zero, whereas at the extremities the force pulling the mass back in is a maximum, therefore the corresponding acceleration must also be also be a maximum.

The other point which might cause confusion is velocity being zero at the extreme positions, but remember we came across this before; when an object is thrown up its instantaneous velocity is zero at the very top, (even though its acceleration is not zero).

“But I thought ω represented angular velocity?”ω is the same symbol as was used in the last chapter (Circular Motion) to represent angular velocity, however it does not represent angular velocity here. It’s not just coincidence that ω turns up in both chapters; they are connected, but the relationship is a little complicated.

In the chapter on Circular Motion we made use of the formula v = distance/time and applied it to a satellite undergoing a full cycle where v = 2πr /T.We also had v = rω, so we can re-write the previous equation as rω = 2π r/T. Cancel r’s to get ω = 2π /T.Cross-multiply to get T = 2π/ω.

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Formulaea = accelerationA = Amplitudex = distance from equilibrium position = constantv = velocity at any time tT = periodic time (time for one complete oscillation)

Know the following equations

a = 2xBasic equation which defines simple harmonic motion

v2 = 2(A2 – x2)

x = A Sin t (if the clock starts (t = o) as the particle is passing the equilibrium position)

x = A Cos t (if the clock starts when the particle is at the extreme position – but x still represents displacement from the equilibrium position, not the extreme point)

amax = 2A Can you see how we get this from the first equation above?

vmax = A Can you see how we get this from the second equation above?

T=1f T=2 π

❑ Where f = frequency = the number of oscillations per second.

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TIP: Make sure that your calculator is in radians mode

Manipulation of equations

Notes The time to go through one complete oscillation is the period T. The time to go from one extreme to the other corresponds to the time to go through half an oscillation,

and this equals T/2. The time to go from the release point to the equilibrium position corresponds to the time to go through

one amplitude, and this equals T/4.

However at any other sub-intervals the distance travelled is not directly proportional to the time, and this is why we revert to the equation: x = A Sin t etc.Also with these equations, x always corresponds to the displacement between the given position and the equilibrium position. This can cause confusion if the object is released from an extreme and the formula: x =A Cost is used.

Know the significance of the phrase ‘when the string is taut’, and ‘when the string is slack’ (when the string is taut the object will be undergoing SHM, but not when it is slack).Quite often the object will only be exhibiting SHM for part of its motion; you must be able to figure out when it is and when it is not exhibiting SHM. It will only be exhibiting SHM when the string is stretched.You should be able to use VUAST for the other situations.

The approach is VERY similar to how you answer vuast-type questions: Write down what you know in terms of the variables Write down all the equations to remind you of your options Identify the equation you need Solve

OMG, I can’t remember the difference between extension and amplitude!!!This is a killer. Picture the situation; a string has got a natural length of 1 m. The string is fixed at one end and there is a mass of 3 kg at the other end.The string is now pulled a further 2 metres and is then released.

The extension is the distance between the object’s Current Position (C.P.) and the end of the natural length of the string (in other words, how much the string is extended by from its natural length).The extension continually changes as the mass moves back.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.). In these situations the equilibrium position corresponds to the point at the end of the string’s natural length. Why? Two reasons (at least); it is the point beyond which the mass will exhibit SHM if released, and secondly it also corresponds to the point of maximum velocity (and zero acceleration) as the mass passes back through.

In the scenario above the extension changes with time but the amplitude is fixed at 2 m throughout the question, regardless of where the mass is.

Let’s go

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2003 (a) A particle is moving with simple harmonic motion of period π seconds about a fixed point o.The maximum speed of the particle is 8 cm/s.(i) Find the amplitude of the motion. (ii) Find the speed of the particle when it is at a distance of 3 cm from o.

1979 (a) A body is moving with simple harmonic motion, of amplitude 5 m. When it is 4 m from the midpoint of its path its speed is 6 m/s. Find its speed when it is 2.5 m from the mid-point.

2012 (a) A particle of mass 0·5 kg is suspended from a fixed point P by a spring which executes simple harmonic motion with amplitude 0·2 m.The period of the motion is 2 seconds.Find (i) the maximum acceleration of the particle(ii) the greatest force exerted by the spring correct to one place of decimals.

2001 (a) A particle moving with simple harmonic motion has speeds of 5 cm/s and 2 cm/s when it is at points 3 cm and 4 cm, respectively, from the centre of the motion. (i) Find the amplitude and the period of the motion. (ii) Find the maximum speed of the particle.

1996 (a) A body of mass 10 kg moves with simple harmonic motion. At a displacement of 0.8 m from the centre of oscillation, the velocity and acceleration of the body are 2 m/s and 20 m/s2 respectively. Find(i) the number of oscillations per second

{for some reason the marking scheme assumed you were asked for the number of oscillations per minute – don’t why they made this mistake}

(ii) the amplitude of motion(iii) the maximum acceleration and hence show that the force to overcome the inertia of the body at the

extremity of the oscillation is 223.6 N.

1993 (a)A particle of mass m moves with simple harmonic motion under the action of a variable force.

If the maximum value of the force is

7 m16 and the amplitude of the motion is 4 m calculate

(i) the period of the oscillation

(ii) the speed of the particle at a time

2 π√7 seconds after passing through the centre of oscillation.

2010 (b)A particle moves with simple harmonic motion of amplitude 0.75 m.The period of the motion is 4 s. Find (i) the maximum speed of the particle

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(ii) the time taken by the particle to move from the position of maximum speed to a position at which its speed is half its maximum value.

1990 (a)(i) A particle starts from rest, and moves with simple harmonic motion of period 6n seconds. (ii) Show that the particle moves from the position of maximum velocity to the position in which the

velocity is half the maximum in n seconds.

2006 (a) A particle moves with simple harmonic motion of period 3π. At time t = 0, the particle passes through the centre of the oscillation. It passes through a point distant 4 m from the centre of motion with a speed of 5 m/s away from the centre.Find, correct to two decimal places, (i) the maximum acceleration of the particle(ii) the time which elapses before it next passes through this point.

1986 (a)A particle moves with simple harmonic motion through two points p and q 1.2 m apart. Its velocity is the same at p as at q. It takes 3 seconds to move from p to q and 3 seconds to move from q to q i.e. passing q the next time. Using a diagram or otherwise find the period of the particle and the amplitude.

1980 (b)A particle is moving in a straight line with simple harmonic motion. When it is at a point p1 of distance 0.8 m from the mean-centre, its speed is 6 m/s and when it is at a point p2 if distance 0.2 m from the end position on the same side of the mean-centre as p1, its magnitude is of magnitude 24 m/s2.

If r is the amplitude of the motion, show that 23= r−0 .2

r2−0.64 and hence find the value of r. Find also the period of the motion and the shortest time taken between p1 and p2 correct to two places of decimals.

1981 (a) {ODDBALL}

Establish the formula T = 2 π √ 1

g for the periodic time of a simple pendulum of length l.The length of a seconds pendulum (T = 2 secs) is altered so as to execute 32 complete oscillations per minute. Calculate the percentage change in length.

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Show that the particle is exhibiting simple harmonic motion

You need to have calculus (differentiation) covered in maths before you tackle this particular section.

Start with x = a sin ωt and differentiate twice.

1983 (a)Define simple harmonic motion in a straight line and show that x = a sin ωt can describe such motion, when x is the distance from a fixed point and a, ω and t have the usual meanings.

1980 (a)A particle is moving in a straight line such that its distance x from a fixed point at time t is given by

x = r cos ωt.Show that the particle is moving with simple harmonic motion.

1998 (a)Define Simple Harmonic Motion. The distance, x, of a particle from a fixed point, o, is given by: x = 7 sin ωt + 24 cos ωt, ω being a constant.(i) Show that the particle is describing simple harmonic motion about o.(ii) Calculate the amplitude of the motion.

1992 (a)If the displacement of a moving particle at any one time t is given by the equation x = 5 Cos ω t + 12 Sin ωt(i) show that the motion is simple harmonic motion.(ii) calculate the amplitude of the motion.

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Questions involving ε

If the clock starts when the particle is at an angle from Equilibrium Position (i.e. in between the equilibrium and the extreme point) then the motion of the particle is described by the equation:

x = A Sin (t + ) where represents the angle at time t = 0

Remember that angles are measured in radians

1974 (a)Define Simple Harmonic Motion in a straight line and show that if the magnitude of the displacement from the equilibrium position after time t is given by x = a sin(ωt + α), where a, ω, α are constants, then the notion is simple harmonic.

2011 (a)The distance, x, of a particle from a fixed point, O, is given by

x=asin ¿ε)where a, ε and ω are positive constants.(i) Show that the motion of the particle is simple harmonic.(ii) A particle moving with simple harmonic motion starts from a point 1 m from the centre of the motion

with a speed of 9.6 m s-1 and an acceleration of 16 m s-2.Calculate a, ω and ε.

2009 (a) The distance, x, of a particle from a fixed point, o, is given by x = a cos (ωt +ε ) where a, ω and ε are constants.(i) Show that the motion of the particle is simple harmonic.(ii) A particle moving with simple harmonic motion starts from a point 5 cm from the centre of the motion

with a speed of 1 cm/s. The period of the motion is 11 seconds. Find the maximum speed of the particle, correct to two decimal places.

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1982 (b)The distance, x, of a particle from a fixed point, o, is given by x = a cos (ωt + α) where a, ω, α are positive constants.(i) Show that the particle is describing simple harmonic motion about o and calculate ω and α if the velocity

v = –2a and x = 3 a5 when t = 0.

(ii) After how many seconds from the start of the motion is x = 0 for the first time? (See Tables P.8. Take pi = 3.142).

1999 (a)

A particle moves with simple harmonic motion of period π2 .

Initially it is 8 cm from the centre of motion and moving away from the centre with a speed of 42 cm/s.Find an equation for the position of the particle in time t seconds.

2004 (b) A particle moves in a straight line such that its displacement from a fixed point o at time t is given by x = a cos(ω t − β) where a, ω and β are positive constants.(i) Show that the motion of the particle is simple harmonic motion.(ii) The period of the motion is 16 seconds. At time t = 4 s, the particle is 12 m from o and 4 s later the

particle is on the other side of o and at a distance of 5 m from o.Find a, ω and β.

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Introduction to Problem-Solving type questions in the x-direction

Hooke’s LawHooke’s Law states that when an object is stretched, the restoring force is directly proportional to the displacement, provided the elastic limit is not exceeded.

We represent this mathematically as follows: F - extension

The minus sign indicates that the restoring force and the extension are in opposite directions.

Or F = - k x

The most common example of Hooke’s Law is a stretched string.

When using the F = -k x expression for Hooke’s Law, x represents the extension, i.e. the distance between the new length and the original (natural) length.

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F = - k (extension)

Wait; I’ve forgotten the difference between extension and amplitude!!!This is a killer. Picture the situation; a string has got a natural length of 1 m. The string is fixed at one end and there is a mass of 3 kg at the other end.The string is now pulled a further 2 metres and is then released.

The extension is the distance between the object’s Current Position (C.P.) and the end of the natural length of the string (in other words, how much the string is extended by from its natural length).The extension continually changes as the mass moves back up.

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.). In these situations the equilibrium position corresponds to the point at the end of the string’s natural length. Why? Two reasons (at least); it is the point beyond which the mass will exhibit SHM if released, and secondly it also corresponds to the point of maximum velocity (and zero acceleration) as the mass passes back through.

In the scenario above the extension changes with time but the amplitude is fixed at 2 m throughout the question, regardless of where the mass is.

To begin with, it doesn’t matter where the object is released from!!! This is because the amplitude only comes into play towards the end of the question.You will usually be told where the particle is released from in the question, but ignore this information completely until after you have demonstrated that the particle is exhibiting S.H.M.

Notes The time to go through one complete oscillation is the period T. The time to go from one extreme to the other corresponds to the time to go through half an oscillation,

and this equals T/2. The time to go from the release point to the equilibrium position corresponds to the time to go through

one amplitude, and this equals T/4.

However at any other sub-intervals the distance travelled is not directly proportional to the time, and this is why we revert to the equation: x = A Sin t etc.Also with these equations, x always corresponds to the displacement between the given position and the equilibrium position. This can cause confusion if the object is released from an extreme and the formula: x =A Cost is used.

Know the significance of the phrase ‘when the string is taut’, and ‘when the string is slack’ (when the string is taut the object will be undergoing SHM, but not when it is slack).Quite often the object will only be exhibiting SHM for part of its motion; you must be able to figure out when it is and when it is not exhibiting SHM. It will only be exhibiting SHM when the string is stretched.You should be able to use VUAST for the other situations.

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Example: 1996 (b) A light elastic string of natural length a and elastic constant k is fixed at one end to a point p on a smooth horizontal table. A particle of mass m is attached to the other end of the string.Initially the particle is held at rest on the table at a distance 2a from p, and is then released.Prove(i) that the particle executes simple harmonic motion while the string is taut.

(ii) that the particle reaches p after ( π

2+1)√ m

k seconds.Solution

Step 1: Show SHM by extending the object a distance x beyond its natural length and invoking Hooke’s Law

Begin with our old friend F = maThe only force here is the restoring force, and from Hooke’s Law this can be expresses as F = - k x

So again F = ma-k x = maa = - (k/m) x

But this is similar in appearance to our general expression for SHM: a = -ω2 x Where k/m is equivalent to ω2, or ω = √k/m

Calculate the period TT = 2/Note the AmplitudeAt this stage we look again at the question to see where the object is released from.The amplitude ‘A’ corresponds to the distance between this release point and the equilibrium position.In this question the amplitude corresponds to a distance a.{So if it’s not enough that small ‘a’ usually represents ‘acceleration’, and capital ‘A’ represents ‘amplitude’, in this question they also decided to let small ‘a’ represent a distance also. I’d say those boys setting the exam paper that year rolled around the floor at that one}

Answer the questionOkay, it seems like a silly thing to say, but really it is only at this stage that you need consider what the question is asking you to do.In effect, the question has now been reduced to a part (a) question where we pick the appropriate equation(s) and solve.Usually we need to find the time which it takes the object to go from its release point to some final position.Usually this involves passing through the equilibrium position and out the other side.We deal with this in two stages:The time to go from the release point to the equilibrium position corresponds to the time to go through one amplitude.As this is one quarter of the total oscillation, the time for this stage corresponds to T/4.

In the example above the mass exhibits SHM as it travels from its release point back to equiblirium.This corresponds to a time of T/4. t1 = T/4

Its speed is now vmax (use vmax = ωA to get a value for this) and it now maintains this speed as it travels between the equilibrium position and the original fixed point.So for this section we use t2 = a/vmax {from time = distance/speed}Total time = t1 + t2

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Horizontal Motion exam questions2005 (b) A light elastic string of natural length a and elastic constant k is fixed at one end to a point o on a smooth horizontal table. A particle of mass m is attached to the other end of the string.Initially the particle is held at rest on the table at a distance 2a from o, and is then released.

Show that the time taken for the particle to reach o is .

1996 (b)A light perfectly elastic string of natural length a and elastic constant k is fastened at one end p to a fixed point of a smooth horizontal table, and a particle of mass m is attached to the other end. The particle is held on the table at a distance 2a from p and then released. Prove(i) that the particle executes simple harmonic motion while the string is taut.

(ii) that the particle reaches p after ( π

2+1)√ m

k seconds.

1992 (b)A particle of mass 2 kg is attached to one end of a light elastic string of natural length 1 m and elastic constant 14 N/m. The other end of the string is fixed to a point A on a smooth horizontal table. The particle is pulled across the table and released from rest at a point C which is a distance 1.5 m from A. If B is a point on AC such that |AB| = 1 m,(i) prove that the particle performs simple harmonic motion when travelling from C to B.(ii) calculate the time taken to travel from C to B.

(iii) prove that the particle then travels for

4√7 s with constant speed.

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The following questions are a bit tricky to tackle at this stage. Best to move on to the next section; ‘Vertical SHM’ and then come back to this after.

1999 (b)

A smooth particle of mass 0.5 kg at rest on a smooth horizontal table is attached to two points p and q, which are 1.2 m apart, by two light elastic strings. The string attached to p has a natural length 0.4 m and elastic constant 75 N/m. The string attached to q has a natural length 0.6 m and elastic constant 50 N/m.(i) Find the equilibrium position.(ii) Prove that if the particle is displaced in the direction p⃗q, through such a distance that neither string goes

slack and is then released, it moves with simple harmonic motion.

1975a, b are two points 4 m apart on a smooth horizontal table and n is the midpoint of [ab]. A particle of mass 0.1 kg is held at n by two elastic strings the other ends of which are attached to a,b respectively. Each of the strings is of natural length 1 m and elastic constant 5 N/m. If the particle is then drawn aside along ab and released from rest when 1.5 m from b, show that it moves with simple harmonic motion. Find the period, and the least time taken to reach a point 1.75 m from b.

1990 (b)The depth of water in a harbour is assumed to rise and fall with time in simple harmonic motion. On a certain day the low tide had a height of 13 m at 12:58 p.m. and the following high tide had a height of 18 m at 6:58 p.m. If a ship requires a depth of 16.5 m of water before it can leave the harbour, find the latest time on that day that the ship can leave the harbour.

1980If a string whose elastic constant is k is stretched a distance x beyond its natural length, show that the work done is ½kx2.

A particle of mass m is on a rough horizontal plane and is connected to a fixed point p in the plane by a light string of elastic constant k. Initially the string is just taut and the particle is projected along the plane directly away from p with initial speed u against a constant resistance F. Find an expression for the distance x travelled by the particle. Noting that the particle will just return to its point of projection if the potential energy at any point is equal to the work done up to that point in overcoming F, show that kmu2 = 8F 2.

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Friction1983(i) Define simple harmonic motion in a straight line and show that x = a sin ωt can describe such motion,

when x is the distance from a fixed point and a, ω and t have the usual meanings.(ii) A particle p, of mass 5 kg, is connected by a light elastic string, of natural length 2 m and elastic constant

140 N/m to a fixed point q on a rough horizontal surface where the coefficient of friction is 1. p is released from rest at a point a where |qa| = 3 m. By considering the forces acting on p when its distance is (2.35 + x) m from q, prove that p moves in simple harmonic motion as long as the string remains taut.

(iii) State the position of the centre, o, of the simple harmonic motion i.e. |qo| and write down the amplitude.

(iv)If the periodic time is assumed to be π√7 calculate the time taken by the particle to travel from a to a

point 2 m from q.

1979 (b)A block rests on a rough platform which moves to and fro horizontally with simple harmonic motion. The amplitude is 0.75 m and the block undergoes 20 complete oscillations per minute. If the block remains at rest relative to the platform throughout the motion, find the least possible value the coefficient of friction can have.

2009 (b) A table moves in a horizontal plane with simple harmonic motion.The table completes N oscillations per minute.Find, in terms of μ and N, the greatest allowable amplitude of the motion if an object placed on the table is not to slip, where μ is the coefficient of friction.

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VERTICAL MOTION

With vertical motion, we have one extra step compared to the previous section.With horizontal motion the Equilibrium Position was taken to be the end of the string, but here it’s a little more complicated. Equilibrium position is the place where the mass comes to rest. At this position the force downwards equals the force upwards, and we use this to calculate d.Now add this distance to the natural length to get Equilibrium Position.

OMG, I can’t remember the difference between extension and amplitude!!!

The extension is the distance between the object’s Current Position (C.P.) and the end of the natural length of the string (in other words, how much the string is extended by from its natural length).

The Amplitude is the distance from the Release Point (R.P.) to the Equilibrium Position (E.P.). The extension changes with time but the amplitude is fixed throughout the question, regardless of where the mass is.

To begin with, it doesn’t matter where the object is released from!!! This is because the amplitude only comes into play towards the end of the question.You will usually be told where the particle is released from in the question, but ignore this information completely until after you have first demonstrated that the particle is exhibiting S.H.M.

Notes The time to go through one complete oscillation is the period T. The time to go from one extreme to the other corresponds to the time to go through half an oscillation,

and this equals T/2. The time to go from the release point to the equilibrium position corresponds to the time to go through

one amplitude, and this equals T/4.

However at any other sub-intervals the distance travelled is not directly proportional to the time, and this is why we revert to the equation: x = A Sint etc.Also with these equations, x always corresponds to the displacement between the given position and the equilibrium position. This can cause confusion if the object is released from an extreme and the formula: x =A Cost is used.

Know the significance of the phrase ‘when the string is taut’, and ‘when the string is slack’ (when the string is taut the object will be undergoing SHM, but not when it is slack).Quite often the object will only be exhibiting SHM for part of its motion; you must be able to figure out when it is and when it is not exhibiting SHM. It will only be exhibiting SHM when the string is stretched.

You should be able to use VUAST for the other situations.So to summarise

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1. Find the equilibrium position E.P.At the point of equilibrium (EP) the force down = the force up.The force down is just the force of gravity: mgThe force up is due to the stretched string so we invoke Hooke’s Law to describe it: F = k (extension).So we have mg = k(ext).

If we refer to this extension as ‘d’ it avoids confusion with ‘x’ which we should be using in the next step.

Usually you will be given k and you will have to find the extension (although sometimes you will be told what the extension is and gots to find k).

2. Show that the object exhibits SHM (and find )As before, extend the object a distance x from equilibrium.The overall extension is now (d+x)Now let Fsmall – Fbig = ma

Again the force down is mg and the force up is k (ext), but now the extension is (d + x).In this case mg is the smaller of the two forces (if we extended the spring and released it, it would accelerate upwards).The reason we go with this rather than with Fbig – Fsmall is so that we get a minus in our expression at the end.

Multiply out to get mg – kd – kx = ma.Usually the mg and kd terms cancel out to give a = (k/m) x, as beforeThe formula is rearranged to give a = - (constant) x. This is similar in appearance to our general expression for SHM: a = -ω2xThe constant therefore represents 2.

Now proceed to the rest of the question.We know so we quickly get a value for T.We now (and only now) look at where the object was released and use this to calculate the amplitude A.Then see what they want us to do.

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Step two: mg - k(d+x) = ma

Step one: mg = k(d)

Recap

Before we start doing the exam questions you need to make sure that you can answer all the following questions:

1. What is Step 1: Find ______________________________________________________

2. How do we do this mathematically? __________________________________________

3. What is Step 2: __________________________________________________________

4. How do we do this? _________________________ and then mathematically __________________

5. Why don’t we refer to the release point at this stage? _____________________________

6. Extension refers to the distance between _______________________________________

7. ‘x’ refers to the distance between _____________________________________________

8. Amplitude refers to the distance between ______________________________________

9. When do we use x = A Sin t and when do we use x = A Cos t?

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Exam Questions

1991 (b)A particle of mass 3 kg is suspended by a light string which is found to extend by 20 cm when the particle is at rest. If the upper end of the string is held firm and the particle is pulled down slightly and then released, find the period of the resultant motion.

2007 (a) A particle of mass m kg is suspended from a fixed point p by a light elastic string.The extension of the string is d when the particle is in equilibrium.The particle is then displaced vertically from the equilibrium position a distance not greater than d and is then released from rest.(i) Show that the motion of the particle is simple harmonic.(ii) Find, in terms of d, the period of the motion.

1974 (b)A light flexible elastic string pq, of natural length 1 m and elastic constant 245 N/m, has one end p tied to a fixed point and has a particle of mass 5 kg attached to the end q. The particle is held 1 m below p and is then released from rest to fall under gravity. By considering the forces acting on the particle when it has fallen a distance (0·2 + x) m, show that it moves with simple harmonic motion and that its acceleration is zero when x = 0. Find the time taken to fall a distance 0·3 m.

2016 (b)A particle P of mass 2 kg is hanging from one end of a light elastic string, of natural length 1 m and elastic constant 98 N m–1. The other end of the string is attached to a fixed point A.The particle is now pulled down to a point Q which is 0·4 m vertically below the equilibrium position and released from rest.(i) Prove that, while the string is taut, P moves with simple harmonic motion.(ii) Find the speed of P when the string first becomes slack (no longer taut).(iii) Find the time taken, from release, for P to reach the highest point in its motion.

2008 (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string which hangs vertically. The elastic constant of the string is 500 N/m.The mass is pulled down a vertical distance of 20 cm from the equilibrium position and is then released from rest.(i) Show that the particle moves with simple harmonic motion.(ii) Find the speed and acceleration of the mass 0.1 seconds after it is released from rest.

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2015 (a)A loaded test-tube of total mass m floats in water and is in equilibrium when a length d is submerged, as shown.The upward force exerted by the water on the test-tube is F.(i) Given that F is directly proportional to the submerged length, find the constant of

proportionality in terms of d, m and g.(ii) The test-tube is now pushed down a small amount and then released.

Show that it will oscillate with simple harmonic motion, and find the period of the motion.

2013 (a) A rectangular block of wood of mass 20 kg and height 2 m floats in a liquid. The block experiences an upward force of 400d N, where d is the depth, in metres, of the bottom of the block below the surface. Find(i) value of d when the block is in equilibrium(ii) the period of the motion of the block if it is pushed down 0·3 m from the

equilibrium position and then released.

1998 (b)An elastic string of natural length one metre is extended 20 cm by a particle attached to its end and hanging freely. The particle is then pulled down a further distance of 40 cm and released.Show that the particle moves with simple harmonic motion when the string is taut.Find the height above the equilibrium position to which the particle will rise.

2001 (b) A particle of mass m kg is suspended from a fixed point p by a light elastic string of natural length l and

elastic constant4mg

l .(i) Find the distance of the equilibrium position from the point p, in terms of l.

(ii) The particle is pulled down until it is at a distance 74l

vertically below p and is then released from rest. Find the time taken, in terms of l, for the string to go slack.

1987Define simple harmonic motion.A particle of mass m is suspended from a fixed point p by a light extensible string of natural length d and

elastic constant

49 md .

It is pulled down a distance

8d5 below p and is then released from rest.

(i) Show that the particle moves with simple harmonic motion as long as the string remains taut.(ii) Find in terms of d, when the string becomes slack for the first time.

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2014 (a) A particle of mass 5 kg is suspended from a fixed point by a light elastic string. The mass is set vibrating vertically with simple harmonic motion, making 4 oscillations per second.(i) Find the constant of elasticity of the string.(ii) The mass is now removed from the string.

Find the force required to stretch the string through 1.5 cm from its natural length.

1985Define simple harmonic motion.A body of mass 0.25 kg hangs from a spiral spring. When pulled down 10 cm below its equilibrium position and released, it vibrates with simple harmonic motion of period 2 seconds.(i) Find its velocity as it passes through the equilibrium position.(ii) What is the shortest time taken to travel from a point 2 cm below the position to a point 2 cm above the

equilibrium position?(iii) Find the elastic constant of the spring.(iv)By how much will the spring shorten when the body is removed?

******************************************************************************This ends the first run-through on this topic – all further questions are a bit trickier and should be left

for the next time the topic is being covered.******************************************************************************

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1989Define Simple Harmonic Motion.A mass of 4 kg suspended by a light spiral extends it 8 cm when in equilibrium. A second mass of 2 kg is attached to the first without moving it and the combined mass is then released from rest.(i) Prove that the motion is simple harmonic.(ii) Find the periodic time of the ensuing motion.(iii) Find the maximum velocity of the resulting motion.

1993 (b)A light elastic string, of elastic constant 48mg/l and natural length l has one end attached to a fixed point. Two particles of masses 3m and 2m are attached to the other end and the system hangs in equilibrium. If the 2m mass falls off (i) prove that the 3m mass will move with simple harmonic motion of period (ii) find the amplitude of the motion.

2000 (b) A particle of mass 0.3 kg is attached to the midpoint of a light elastic string of natural length 1 m and elastic constant k. The string is then stretched between two points a and b. The point a is 2 m vertically above b.(i) Find the extensions of the two parts of the string, in terms of k, when the system is in equilibrium(ii) Find the minimum value of k which will ensure that the lower part of the string is taut(iii) Find the period of small oscillations, in terms of k, when the particle is displaced vertically.

(Assume both parts of the string remain taut.)

1976Define simple harmonic motion.A particle of mass 2 kg is attached to the ends of two elastic strings, each of natural length 1 m and elastic constant 49 N/m. The other ends of the two strings are attached to two fixed points a and b in the same vertical line where a is 4 m above b. The particle is released from rest from the midpoint of ab. By considering the forces acting on the particle when it is x metres from a, where 2 < x < 2.4, show that it is moving with simple harmonic motion. Find the least time taken for the particle to reach the point x = 2.3, and find its speed there.

1997 no.7 (a)A particle of weight W is attached to two inextensible strings each of length 26 cm. The other ends of the strings are attached to two points on the same horizontal level 48 cm apart. Find the tension in each of the strings in terms of W.The strings are replaced by two light elastic strings, each of natural length 26 cm and elastic constant k, attached in the same way. The position of equilibrium of the particle is now 8 cm below its previous position of equilibrium. Find the value of k in terms of W.

1986 (b)

An elastic string of natural length 2a is stretched between two points which are 2√3a apart. A particle of weight W is attached to its midpoint and hangs in equilibrium with the string inclined at an angle of 300 to the horizontal.Show that the modulus of elasticity of the string is W.

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π2 √ l

g

1994A scale pan is suspended from a fixed point P by a light elastic spring. A particle Q of mass 0.2 kg is attached to the pan with glue. The pan is pulled down from its equilibrium position and set in motion.

Given that the motion of Q is simple harmonic, with period

π6 seconds and that the maximum and

minimum distances of Q below P are 1.5 m and 0.9 m, respectively, calculate(i) the maximum speed of Q.(ii) the maximum force that the glue has to exert on Q.(iii) the length of the spring, when, in the absence of glue, Q would leave the pan.

1991 (a)A light elastic string of natural length 1.2 m is found to extend by 20 cm when a small mass is gently attached to one end, the other end being fixed. The mass is now made to describe a horizontal circle with angular velocity rad/s. Find an expression for the extension of the string in terms of .

1979 (b)A conical pendulum consists of a light elastic string with a mass m attached to it which is rotating with uniform angular velocity ω. The natural length of the string is l and its elastic constant is k, i.e. a force k produces unit extension. The extended length of the string is l’ and it makes an angle θ with the vertical.

Prove that k(l’ – l ) = ml’ω2 and that cos θ =

glω2

−mgkl

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Guide to answering the Exam Questions2009 (a)(i) Would be straightforward if not for the term. Either use your knowledge of differentiation or else just

learn this off because it does come up o am go h-am)(ii) Straightforward if you choose the correct equation. Answer: v = 3.03 cm/s

2009 (b)This was quite simply an absolutely horrendous question (and it doesn’t help when letters are used instead of numbers). You get an expression for the maximum friction force (fair enough; F = μR) and a separate expression for the motion involving the “N oscillations per minute” information. I can see where the value for comes from, but I would have thought that the F = mr2 expression referred to circulation motion and not SHM. Presumably I would be wrong. Once you get an expression for both these terms simply equate to get r = amplitude = (893.65 μ)/N2

2008 (a)(i) Straightforward. Ans: a = - 100 x(ii) Straightforward. Ans: a = 10.8 m s-2

2007 (a)(i) Straightforward. Ans: a = - (k/m) x(ii) Straightforward. Ans: T = 2π (d/g)

2006 (a)(i) Straightforward. Ans: amax = 3.78 m s-2.(ii) Straightforward. Work out the time it takes to go from equilibrium to the point 4 metres out, then

take this away from T/4 (the time it takes to go from equilibrium to the extreme) and multiply this by two to get the time to go out and back in.

Ans: t = 3.24 seconds.

2005 (b)Nice question. Find as normal. The time to reach o can then be broken into two stages: time to travel a distance a from the release point. The string is stretched for all of this so we need to find the time (which will be T/4) and use SHM formula to find the velocity at the end of this stage. From here on the string is no longer taut, so the particle is simply moving in a horizontal direction with whatever velocity it had when it first got to that point.

2004 (b)(i) Straightforward, but only if you had seen it before.

x = a cos (ω t − β), d x/dt = - a ω sin (ω t − β), d2 x/dt2 = a ω2 cos (ω t − β ). Can you say why?(ii) First find ω, then get two equations with A and β, then solve.

Ans: ω = π/8, β = 1.176 radians, A = 13

2003 (a)(i) Straightforward. Ans: A = 4 cm, (ii) Straightforward. Ans: v = 27 cm/sec.

2001 (a)(i) Straightforward. Ans: A = 4.16, T = 3.63 (ii) Straightforward. Ans: vmax = 213 cm/sec.

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2001 (b)(i) Straightforward. Distance from p = 5l/4.(ii) Time taken is in two stages; the first is from the release point to equilibrium position which is

equivalent to T/4. The second stage is from equilibrium position to a distance of l/4 above equilibrium position (above this point the string will be slack.

Ans: T = π/3(l/g)

2000 (b)This is a tricky SHM question, but at least it is broken up into parts.

(i) Straightforward. x1 = (k + 0.3g)/2k and x2 = (k - 0.3g)/2k.(ii) Slightly tricky. If the lower part of the string is taut then there must be an extension, i.e. x2 0. Use

this expression to get a minimum value for k as 0.3g.(iii) Straightforward, if testing due to the expressions being slightly complex. T = π(0.6/k)

1999 (a)A little tricky in that there is an term in there because the particle is initially 8 cm from the centre but if you’ve dealt with this before then it should be fine.Answer: x = 66 sin(4t + 1.4)

1999 (b)(i) Straightforward; equilibrium position is 0.48 m from p.(ii) Straightforward.

1998 (a)Very unusual (and therefore imo rather nasty) to get asked a definition, but physics students should know it off by heart at least.This is one of those questions which is very easy if you have seen it before, but if you haven’t . . .To show that the particle is describing SHM, simply treat the two components (7sin t and 24 cos t) separately and proceed as normal.To calculate the amplitude simply look at the amplitude of both components (7 and 24) and get the magnitude using Pythagoras.Answer: Amplitude = 25

1998 (b)Straightforward vertical SHM question. Note that usually you are given k and need to calculate equilibrium position, but in this case you are actually given equilibrium position and need to calculate k. There are two stages involved here;

1997 (a)No. 7

First up, it was unusual to find a SHM question as part (a) on No.7, which is always statics, but there you go.As a SHM question it was fine.

(i) This was very straightforward and remarkably similar to No. 6 part (i) of the same year. No SHM for this part. Forces up = forces down and forces left = forces right. You also needed a little trigonometry to figure out the angle at the bottom.Answer: T = 13W/10

(ii) Straightforward.Answer: k = 125W/6

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1996 (a)All straightforward, just pick the correct equation.No. of oscillations per second = 5/2π Amplitude = 0.89 mMax acceleration = 258 rads/sec

1996 (b)Straightforward once you establish that the particle is undergoing SHM on its way back until it reaches the natural length (a) away from p, at which stage it travels at constant velocity (this velocity corresponds to the max velocity of the SHM stage).

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Summary of key points

The extension of a spring or a string is the distance between the object’s Current Position and the end of the natural length of the string (in other words, how much the string is extended beyond its natural length).

The Amplitude is the distance between the Release Point (R.P.) and the Equilibrium Position (E.P.).

When do you first need to note the release point?After proving that the motion of the particle is simple harmonic.

Equilibrium Position

How do you find equilibrium position for a string in the horizontal direction?Equilibrium position is at the end of the string.

In general, what concept do we use to find equilibrium position for a string in the vertical direction?Force down equals force up.

What will the equation look like (for a string in the vertical direction)?mg = k {ext}

Why should we use the letter ‘d’ to represent extension for this stage?Because we will use ‘x’ to represent distance from E.P. for the next stage.

To show SHM

Complete the following: “To show SHM, extend ___________________________“Extend the object a distance x beyond E.P. and use Fsmall – Fbig = ma”

The first line of the solution to show SHM in the horizontal direction (one string, no friction) will be:– k{ext} = ma

The first line of the solution to show SHM in the vertical direction (assume only one string) will be:mg – k{ext} = ma

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Test yourself

1. The extension of a spring or a string is the distance between

____________________________________________________________________________________

2. The Amplitude is the distance between

____________________________________________________________________________________

3. In equation a = -2 x, x represents the distance between

____________________________________________________________________________________

4. When do you first need to note the release point?____________________________________________________________________________________

5. How do you find equilibrium position for a string in the horizontal direction?

____________________________________________________________________________________

6. In general, what concept do we use to find equilibrium position for a string in the vertical direction?

____________________________________________________________________________________

7. Why should we use the letter ‘d’ to represent extension for this stage?

____________________________________________________________________________________

8. Complete the following: “To show SHM, extend

___________________________________________________________________________________”

9. What will the first line of the solution (to show SHM) look like?

____________________________________________________________________________________

10. The first line of the solution to show SHM in the horizontal direction (one string, no friction) will be:

________________________________________________

11. The first line of the solution to show SHM in the vertical direction (assume only one string) will be:

________________________________________________

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