Simple harmonic motion

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Simple Harmonic Motion

Simple Harmonic Motion- By Aditya Abeysinghe


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Consider an object moving round a circle with center O and rotating with a uniform angular speed ω.

Acceleration at A is ω2r, and this acceleration is directed along the radius AO.


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Hence the acceleration of P towards O is w2rCosAOC,Which is also equal to w2rSinθBut r Sinθ = PO = y say.Therefore, acceleration of P towards O is w2ySince w2 is a constant,Acceleration of P towards O α distance of P from O.Since the acceleration of the object is always directed towards the center, acceleration should be expressed as,Acceleration towards O = - ω2y.The minus sign indicates that the object begins to slow down its motion as it passes the center of its path.


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Thus, Simple Harmonic Motion can be expressed as follows:

It is the motion of any object whose acceleration is always directly proportional to the distance from its center of path and the acceleration is always directed towards the center.


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The relationship between Simple Harmonic Motion and Circular motionConsider an object A rotating round its center of path O.

B

C

amplitude

Period (one cycle)


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The B and C points show the corresponding points where the circular motion is equivalent to the wave. Since the circular motion occurs continuously, it can be concluded that the corresponding wave pattern also equals periodically.Thus Simple Harmonic Motion is the same concept as moving an object in circular, continuous motion.


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Amplitude of MotionConsider an object attached to a spring as shown below.

X = 0 X = XmaxX = Xmin

Direction of motion

amplitudeamplitude


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When you release the mass, the spring will exert a force, pushing the mass back until the object reaches the position x=xmax . This position where the object has reached its maximum is called its

amplitude.

Theoretically,Xmin = - Xmax .


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Hooke’s LawLaw: The force exerted on a body by an external source is directly proportional to the mean displacement the object has displaced from its mean position.Consider a spring-mass system as shown below.

X = 0

F

X – displacement from mean position


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According to Hooke’s law,F α x . (F-Force, X- Mean displacement)However, theoretically the force on the object is always acting so that the direction of force always opposes the direction of displacement. For example, when the spring travels to its amplitude from its mean position, x increases while F decreases. Similarly when the object returns to its original position the force exerted on the object gradually increases while the displacement from mean position decreases.Thus, theoretically,F α – X.


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To eliminate the proportionate sign, we use a constant k, which is called the spring constant (since it depends on the spring used).

Thus, F= -kX.Since, F=ma, by Newton’s 2nd law of motion, the direction of F is the direction of acceleration. Thus, like acceleration, the force that acts on an object in simple harmonic motion is always directed towards the center of its path.

Since, this force acts towards equilibrium, it is called a restoring force.


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Displacement of an object in Simple Harmonic Motion

Sin θ = x/AX = A Sinθθ = ωt

X = A Sin ωtHowever, if the object when released (at time=0) is not at the center of its path, then the displacement of the object is not accurate to be displaced as above.Thus for accuracy, the displacement equation is generally expressed as,

X = A Sin (ωt + φ), where φ is the phase lag.

X

X

Aθ

t = 0

t = tω


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Period in Simple Harmonic Motion

Since the simple harmonic motion is analogous to circular motion the period of an object in simple harmonic motion is the same as that in circular motion.Thus, the period is the,The time taken by an object to complete one total rotation.Thus, this can be graphically expressed as,


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Since θ for one complete revolution is 2π,And that ω = θ/t ,ω = 2π/t or the period of motion,

T = 2π / ω.

However, f = 1/ T, where f is the frequency of motion, or the no. of rotations per unit time,

It is evident that, f = ω / 2π or ω = 2πf.

Relationship between velocity and displacement

V = V’ CosθBut, Cos θ = √(A2 – x2)

And, V’= ωr Since r = A or the amplitude,V’ = ωAThus, from the above derivations,V= ωA Cosθ or V= ωA √(A2 – x2) Thus,

V= ω √(A2 – x2)Simple Harmonic Motion- By Aditya

Abeysinghe15

A

X

X

A θ

ω

√(A2- X2)V’

θ

V


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Potential energy of a springThe potential energy of a spring, also sometimes called the elastic potential energy can be calculated as follows:

U = - ∫ F dx.However, we know that F = -kX.

Thus, U = - ∫ (-kx) dx = ∫ kx dx

= k ∫x dx = k { x2/2}

Thus, U= ½ kx2


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Oscillating Systems- Spring and Mass

From Hooke’s law, From Newton’s second F = -kX law of motion, F = ma

X

F


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Therefore, ma = - kxThus, a = - (k/m) x.However, we know that for any object experiencing simple harmonic motion,a= -ω2x. Therefore, -ω2x = - (k/m) x.

Thus, ω2 = k/m. Therefore, the period of this motion, T,T = 2π/ω = 2π/ {√(k/m)}

T = 2π √m/k


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Principle of conservation of energy in simple harmonic

motionBy applying principle of conservation of energy,

Kinetic energy = Potential energy at any position at any position

Thus, ½ mr2ω = ½ kx2

Oscillation of springsOscillations of springs can be described under two categories based on the mode of oscillation as,1. Horizontal oscillations2. Vertical oscillations

The main difference between the two types of oscillations is that in vertical oscillations the gravitational potential energy too contributes towards(in addition to elastic potential energy and kinetic energy, as in horizontal oscillations) the total energy at any moment in the motion.


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Horizontal oscillation of springs


At xmin and at xmax the object gains the maximum potential energy, as the object has reached its amplitude of motion.The object gains the maximum velocity, hence the maximum kinetic energy as it passes the mean position or the center of its path

The total energy of the spring-mass composite = elastic potential energy + kinetic energy. The total energy is conserved for the system or the system obeys conservation of mechanical energy

Xmin XmaxX = 0

Vertical oscillation of springs

Simple Harmonic Motion- By Aditya Abeysinghe22

Similar to that in horizontal oscillation , the object gains maximum elastic potential energy at its amplitude and maximum kinetic energy at its center of path.

However, total energy of the spring-mass composite = elastic potential energy + gravitational potential energy + kinetic energy


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Equilibrium PositionBecause the mass exerts a gravitational force to stretch the spring downward, the equilibrium position is not the position where x = 0, but, x = -h, where h is the vertical displacement of the spring due to gravity.

The equilibrium position is the place where the net force acting on the object is zero

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Restoring force acting on the spring = -khThe gravitational force acting on the object = mgHowever, for the object to be stationary, by applying F = ma for the composite a = o and thus the restoring force = gravitational force.

Thus, -kh = mg or h = -(g/k)m.Thus, it should be noted that hα m. (higher masses drag the spring further towards gravity) Simple Harmonic Motion- By Aditya Abeysinghe

Initial Equilibrium Position(Position without weight)

Final Equilibrium Position(Position gained due to weight mg)


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Coupled Spring Systems1. Springs in seriesConsider two springs with spring constants k1 and k2 connected to a mass m. When m is displaced by a distance x, and if the displacements made by the two springs are x1 and x2 respectively,

Then x1 + x2 = x.


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If the restoring force is F, thenF = -k 1x1 = -k2 x2. Thus, x1 = -F/k 1 and x2= -F/ k2

Since, x1 + x2 = x , it can be derived that

X = {-F/k 1 } + {-F/ k2 }.

By applying Hooke’s law to the spring composite, we can write it as,

F = -keff X , and hence X = -F/keff. (keff – keffective)

By substituting this to the above derivation, we find that,

keff = k 1 k2 /(k 1 + k2)And period of oscillation,T,

T = 2π √ m { (k 1 + k2)/ k 1 k2 }


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2. Springs in parallel In this case if m is displaced by a distance of x, both springs display by x. However, the force exerted by the two springs on m on extension are unequal.

For first spring, For second spring,F1 = -k1x F2 = -k2 x


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Since the total force on m is equal to the summation of the individual forces exerted by the two springs,F = F1+ F2

Thus, -keff x = (-k1x )+ (-k2 x)

Therefore,

keff = k1 + k2

The period of motion, T,

T = 2π √ m / (k1 + k2)


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3. Masses connected between two strings

If the body is displaced, one of the strings get extended while the other gets compressed.Thus, the total force on the object is the sum of the two restoring forces exerted by the two springs.

Thus, F = F1+ F2

As in the case of springs in parallel, keff = k1 + k2


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Pendulums

F = ma = mg SinθHowever, for small angles, Sinθ = θ, and, θ = s/l Thus, F = mgs/lTherefore, ma = mgs/l.Thus, a = gs/l.Since, the force is in the opposite direction of the displacement. It should be written as ,a = -gs/l or a = -(g/l) sThis is in the form a = -ω2x and the pendulum hence shows a Simple Harmonic

Motion, where, ω = g/l.

F’

mg Cosθmg Sinθ

θ

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Period of motion-

T = 2π/ω . Therefore, T = 2π √(l/g) Energy-The mechanical energy of a pendulum is a conserved quantity.


θ

θ = θmax θ = 0θ = θmin

Velocity = 0 Velocity= max Velocity = 0

Potential energy = max

Potential energy = min

Potential energy = max

Restoring force = max

Restoring force = 0 Restoring force = max

Tension = min Tension = max Tension = min

Kinetic energy = 0 Kinetic energy = max Kinetic energy = 0


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Velocity-

X = L CosθTherefore, h = L - L CosθFrom the conservation of energy,Total initial energy = Total final energy½ mv2 = mgL (1- Cos θ )

Thus, V = √{2gL (1- Cos θ )}

V α LV α θ

θL

X

h

V

Simple harmonic motion

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