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JURONG JUNIOR COLLEGE PHYSICS DEPARMENT

TYS Solutions Oscillations

Qn Ans Suggested solution 1 C a = -ωx2

=> General shape of a-t graph is a mirror image of the general shape of x-t graph. 2 B For SHM : a = -ω2x a ∝ -x

3 D 2

1 22 2 2= rad s10 10

a x xTπ π πω ω− ⎛ ⎞= ⇒ = − = −⎜ ⎟

⎝ ⎠

4 A Angular velocity ω = 2π/T = 2π/0.063 = 100 rad s-1 Maximum velocity vo = ωxo where xo is the amplitude of oscillation.

Hence xo = voω = (3.0)/(100) = 0.030 m. 5 A xo = 5.0 cm

a = −ω2x = −22

Tπ⎛ ⎞

⎜ ⎟⎝ ⎠

x 10 = −22

Tπ⎛ ⎞

⎜ ⎟⎝ ⎠

(−5 × 10−2) T = 0.44 s

6 C Angular frequency

= 2π x frequency

= 2π x no. of oscillations per sec

= 2π n rad s-1

7 A F-r graph => F proportional to -r,

i.e. force is directed towards fixed point

=> Ep will increase as particle is moved away from fixed point (against F). 8 B 2

022

02 2

21

21 xfmxmE )( πω ==

= 22 0050402008021 ).())(.( π

=6.3 x 10-3 J 9 N02/II/4

(a) 1. The amplitude of the curve is not constant but decreasing. 2. The curve does not start from zero displacement as in a sine curve. Points to note: Common error in referring to decreasing displacement, rather than decreasing amplitude.

(b)(i) From Fig. 4.2, period T = 1.5 s

Angular frequency, ω = 2Tπ = 2

1.5π = 4.19 rad s−1

Points to note: Common error in the unit of angular frequency, given as s−1 rather than rad s−1.

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Qn Ans Suggested solution (b)(ii

) At t = 1.5 s, acceleration = −ω2 x = −(4.19)2 (2.2) = −38.6 cm s−2

(c) Sketch graph (- - -) with the same frequency but increasingly lower amplitude due to greater damping.

Points to note: Should draw accurate sketches with attention being paid to the relevant features. Marks were lost through careless drawing. It was expected that the curve would show amplitude decreasing at a faster rate but with the same, or slightly longer, period. Common error in the curves being drawn with variable half-period and with later peaks having an equal or greater amplitude than earlier ones, and did not draw on Fig. 4.2 despite being told. Those sketches with insufficient detail were not given any credit.

10 N03/II/4

(a) In s.h.m., the particle has an acceleration which is directly proportional to its displacement about an equilibrium position. The acceleration of the particle is always in a direction towards the equilibrium position of the particle.

(b)(i)

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Qn Ans Suggested solution (ii)1.

Frequency = 1/Period = 1

1.5= 0.67 Hz

2. 1 square represents 0.01 cm. There are 72 squares in the shaded region. Therefore, Amplitude = 0.72 cm

(iii)

11 N04/II/1 (b)(i)

1. Force acting on base of tube due to water (Upthrust)

3 3

( )(9.8 10 )(0.055)(1.3 10 )

pAgh Aρ

−

==

= × ×= 0.70 N (2 sf)

2. Vertically upwards. 3. tube is floating, i.e. at equilibrium

⇒ weight of tube = upthrust exerted by water (weight of displaced water) = 0.70 N Points to Note: Some students may make mistakes by giving the unit as kilogram instead of newton.

(ii)1. From Fig 1.3, amplitude of oscillations, 7.0 4.0ox = − = 1.5 cm

2. period of oscillations, T = 0.30 s

frequency of oscillations, 1 1 =

0.30f

T= = 3.3 Hz

3. The tube is oscillating with SHM, i.e. 2 2

ov x xω= ± − Max speed occurs at amplitude = 0,

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Qn Ans Suggested solution

(2 )

2 (3.33)(0.015)

o o

o

v xf x

ωπ

π

==

= = -10.31 m s

Points to Note: Common mistakes include giving peak-to-peak value as amplitude

12 N06/II/3 3(a)(

i) The graph is a straight line with negative gradient passing through the origin. This means that the acceleration of the particle is proportional to its displacement but in the opposite direction. Hence the motion is simple harmonic. [Points to note: Common error to be contented to give a definition as to what is meant by simple harmonic motion but not link features of the graph to the definition.]

(ii) SHM equation: 2a xω= −

Gradient of graph ( ) ( )

( ) ( )6

3 3

3400 34008.5 10

0.4 10 0.4 10−

− −

− −= = − ×

− × − × rad2 s-2

Hence −ω2 = −8.5 × 10-6 rad2 s-2 ω = 2920 rad s-1 2πf = 2920 rad s-1 f =460 Hz

(b)(i) The kinetic energy Ek of a particle in simple harmonic motion is ( )2

2 2 21 12 2kE mv m A xω= = − .

Maximum kinetic energy EMAX occurs when x = 0, hence 2 212MAXE m Aω= .

Since the magnitude of the gradient of the line is G, and G = ω2, 212MAXE mGA= .

[Points to note: Common error in that the steps in the argument (derivation) were frequently omitted. For example, having quoted a general expression for kinetic energy (½mv2), the next line would be the maximum kinetic energy, in terms of A and ω, for an oscillating body.]

(ii) ( ) ( ) ( )22 3 6 3 31 12 2 2.5 10 8.5 10 0.40 10 1.7 10MAXE mGA − − −= = × × × = × J

[Points to note: Common careless mistakes (e.g. using A rather than A2).]

13 N08/II/2

(a) The motion of a body is simple harmonic if the acceleration a is proportional to, and in the opposite direction to, the displacement from the equilibrium position x, i.e . a ∝ −x The given a-x graph is a straight line through the origin, with a negative gradient, a ∝ −x, i.e. oscillations of plate is simple harmonic. Points to note: After definition of simple harmonic motion, a common error is not to go on to explain how the graph showed that the oscillations followed the given definition. Another error is not to state that the graph was a straight line passing through the origin and link this with the definition of acceleration being proportional to the displacement. Should link the negative gradient with that part of the definition that refers to the direction of the acceleration being opposite to the displacement. Common error to suggest the acceleration and displacement were inversely proportional.

(b)(i) Acceleration at which sand first loses contact = g = 9.81 m s-2

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Qn Ans Suggested solution The sand is only acted upon by its weight, i.e. its maximum downwards acceleration is g. When amplitude x0 is sufficiently large, the downward acceleration a0 (∝−x0) at the upper maximum displacement will be equal to the value of g, and the contact force between the sand and the plate will be zero. and the sand will loose contact with the plate. Points to note: Common error to suggest that the resultant force acting on the sand was zero when the sand lost contact with the plate. There are some confusion concerning the resultant force and the reaction force. Common error to suggest that contact force being zero means resultant force is also zero. Should refer to the point where the sand first loses contact as the maximum displacement of the plate and when it was moving in the downward direction. Common error to give the maximum possible acceleration from the graph (20 m s–2).

(ii) From given graph, for a0 = −9.81, amplitude x0 = 3.9 mm

14 N08/III/6 (a) The frequency f of a body undergoing simple harmonic motion (SHM) refers to the number of

oscillations it undergoes per unit time. The angular frequency ω is the circular representation of frequency f and it is the product of the frequency f and the angle of 2π rad, i.e. ω = 2πf. Points to note: Should give the relation ω = 2πf, with explanation.

(b)(i) Loss in GPE = mgh = (0.400)(9.81)(0.200) = 0.785 J

(ii) Spring constant, k = mg

x= (0.400)(9.81)

0.200= 19.62 N m−1

Gain in EPE = ½ kx2 = ½ (19.62)(0.200)2 = 0.392 J

(c) There is no appreciable kinetic energy as the mass is gently lowered to the equilibrium position. The loss in gravitational potential energy is more than the gain in elastic potential energy. The difference is absorbed by the mechanism that lowers the mass gently, exerting an upward force against the gravitational force, thus doing negative work. Points to note: Common error to base answers on the difference being kinetic energy. Should realise that the load was lowered gently.

(d)(i) At the equilibrium position, when the extension of the spring, x = 0.200 m, tension in the spring = weight of mass, i.e. kx = mg At the lowest position, the extension of the spring = 2x, so tension = 2kx resultant force on load = 2kx − mg = kx = (19.62)(0.200) = 3.92 N Points to note: In general, explanation was lacking.

(ii) Angular frequency, ω = k

m= 19.62

0.400= 7.00 rad s−1

OR Resultant force = ma = mω2x 3.92 = (0.400) ω2 (0.200) ω = 7.00 rad s−1 Points to note: Common error to assume that the acceleration is the acceleration of free fall.

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Qn Ans Suggested solution (iii) Maximum speed of mass, vo = ωxo = (7.00)(0.200) = 1.40 m s−1

(e) Complete given table:

GPE / J

EPE / J

KE / J

Total / J

Lowest point 0 1.57 0 1.57 Equilibrium position 0.785 0.392 0.392 1.57 Highest point 1.57 0 0 1.57

Points to note: Common error to complete boxes without reference to their answers in (b) or to the physics of the situation. Should realise that, at the lowest and at the highest points, the kinetic energy would be zero.

(f) Sketch variation with position of energies:

15 N09/III/1

(a) SHM is a periodic motion where its acceleration is proportional and opposite to its displacement from the equilibrium position.

(b)(i) (1)

-12 2 (620) 3900 rad sfω π π= = =

(2) ( )2 2 -2| | 3900 0.21/1000 3190 m soa xω= = =

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Qn Ans Suggested solution (ii)

Points to note: Common error to have no scale drawn on the y-axis or mark just one point, indicating a poor choice of scale.

(c) If the natural frequency of vibration of the cone is within the driving frequency of the sound wave, then resonance will occur, which will cause the amplitude of the sound wave produced to be larger than intended, and hence producing a sound that may be louder than intended. Points to note: Should not paraphrase the question. Common error to refer to the natural frequency being within the range of frequencies, rather than relating the natural frequency to the forcing frequency. Another error is to refer to either an increasing amplitude or an increasing frequency of vibration.

16 N10/III/6

(a) The resultant force acting on an object is the rate of change of momentum of the object. The change in momentum is in the direction of the force.

Points to note: Be reminded that force is a UvectorU quantity and that, where a vector quantity is being defined, Udirection should be taken into considerationU.

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Qn Ans Suggested solution (b)(i)

(ii) Energy stored in spring E = work done in stretching the spring

Since the restoring force in the spring F is directly proportional to its extension x (F = kx),

work done = average force × extension

Therefore E = 2kx x⎛ ⎞

⎜ ⎟⎝ ⎠

= 212 kx

Points to note: Common error to refer to ‘area under the graph’ and then equated L to x without any identification of F with W. Should start with a statement that Uwork doneU is the product of average force and extension.

(c)(i) With 4.0 N weight, length of spring is 9.2 cm.

With further extension of 0.8 cm, total length of spring is 10 cm.

(ii)1. 1. Change in gravitational potential energy = mgh = (4.0)(0.008) = 0.032 J

(ii)2. 2. Increase in elastic potential energy

= average force × extension = 4+5 0.0082

⎛ ⎞⎜ ⎟⎝ ⎠

= 0.036 J

Points to note: Should realize that (x2 2 – x12) is not equal to (x2 – x1)2, and the

substitution into the expression E = ½kx2 involved (42 – 3.22) and not 0.82.

(iii) Increase in elastic potential energy

= change in gravitational potential energy + work done in stretching the 0.8 cm

0.036 = 0.032 + work done in stretching the 0.8 cm

Work done in stretching the 0.8 cm = 0.036 – 0.032 = 0.004 = 4.0 × 10-3 J Points to note: Should involve the difference between the UchangeU in the elastic potential energy of the spring and that of the gravitational potential energy of the mass. Common error to assume that the work done would be equal to the change in the elastic potential energy of the spring.

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Qn Ans Suggested solution (d)(i) 4.0 × 10-3 J

(ii)1. 1. 2 3102 mv 4 0 10−= ×.

-312

4.0 4.0×109.81

20v⎛ ⎞ =⎜ ⎟

⎝ ⎠

v0 = 0.14 m s-1

(ii)2. 2. 0 0v xω=

0.14 = (2πf)(0.008)

f = 2.8 Hz

(e) The frequency of oscillation decreases since the effective mass of the oscillating system increases. Points to note: Common error for explanations to be vague and to consider that the frequency would decrease as a result of damping caused by the increased mass. Another error is to state that m has increased and therefore f has increased. Should state that a spring with mass would Uincrease the effective mass U of the oscillating system.

17 C With critical damping, a system returns to its equilibrium position in the shortest possible

time without oscillating. 18 D A lightly damped oscillation should have decreasing amplitudes and (approximately)

constant period. 19 D Damping causes a decrease in the amplitude xo at all frequencies. The resonant

frequency fo is also slightly reduced.

20 N05/II/4 (a)(i) Forced oscillation is the oscillation produced when a body is subjected to an oscillatory external

force. (ii) Resonance

(b)(i)1.

Maximum linear speed vmax = wx0 = (2πf)(x0) = 2 x π x 12 x 1.6 x 10-2 = 1.21 m s-1

2. Maximum linear acceleration amax = w2x0 = (2πf )2(x0) = (2 x π x 12)2 x 1.6 x 10-2 = 91.0 m s-2

(ii) Time interval = ¼ T = ¼ x (1/12) = 0.0208 s

(c) When the mass of the machine increases, it will take a longer time to complete one oscillation. The period of the oscillation increases and the natural frequency of the driven system will decrease. Thus the graph will shift to the left and the maximum amplitude of oscillation occurs at a frequency that is lower than 12 Hz. Points to Note: Common mistake was not showing that the amplitude as being smaller at all frequencies.

END

TYS 2012 solutions Oscillation - PBworksjjphysics.pbworks.com/w/file/fetch/63726015/11. TYS 2012...

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