CAIE A-Level Physics A2 17 Oscillations Questions

Practise interpreting simple harmonic motion through displacement, velocity and acceleration relationships, energy changes, damping behaviour and resonance response.

Exam points

use SHM parameters, equations and graphs to analyse displacement, velocity, acceleration and phase
calculate and interpret energy interchange and total energy in SHM
analyse damping and resonance using amplitude, natural frequency and response graphs

Covered syllabus points

17.1 Simple harmonic oscillations 17.2 Energy in simple harmonic motion 17.3 Damped and forced oscillations, resonance

[Maximum number: 10]

Fig. 4.1 shows the variation with time t of the height h above the ground of an object of mass 36 kg that is undergoing vertical simple harmonic motion.

Fig. 4.1

(a)

For the oscillations of the object:

[ 6 ]

(i)

determine the amplitude $x_{0}$ , in cm

x_{0}=

[ 1 ]

$x_{0}=8.0 \mathrm{~cm}$

(ii)

show that the angular frequency $\omega$ is $1.6 \mathrm{rad} \mathrm{s}^{-1}$

[ 2 ]

$\omega=2 \pi / T$ $=2 \pi / 4.0=1.6 \mathrm{rad} \mathrm{s}^{-1}$

(iii)

determine the total energy E.

[ 3 ]

$E=1 / 2 m \omega^{2} x_{0}{ }^{2}$ $=1 / 2 \times 36 \times 1.6^{2} \times 0.080^{2}$ $=0.29 \mathrm{~J}$

(b)

On Fig. 4.2, sketch the variation with h of the kinetic energy $E_{\mathrm{K}}$ of the object.

Fig. 4.2

[ 4 ]

dome-shaped curve, starting and ending at $E_{\mathrm{K}}=0$
maximum $E_{\mathrm{K}}$ shown as 0.29 J
position of peak shown at $h=10.0 \mathrm{~cm}$
line intercepts h-axis at $h=2.0 \mathrm{~cm}$ and at $h=18.0 \mathrm{~cm}$

[Maximum number: 11]

A small steel sphere is oscillating vertically on the end of a spring, as shown in Fig. 4.1.

Fig. 4.1

The velocity v of the sphere varies with displacement x from its equilibrium position according to

v= \pm 9.7 \sqrt{\left(11.6-x^{2}\right)}

where v is in $\mathrm{cms}^{-1}$ and x is in cm .

(a)

Calculate the frequency of the oscillations.
frequency = Hz

[ 2 ]

$\omega=2 \pi f$ $\begin{aligned} f =9.7 / 2 \pi =1.5 \mathrm{~Hz} \end{aligned}$

(b)

Show that the amplitude of the oscillations is 3.4 cm .

[ 1 ]

amplitude $=\sqrt{ }(11.6)=3.4 \mathrm{~cm}$

(c)

Calculate the maximum acceleration $a_{0}$ of the sphere.

a_{0}=

$\mathrm{ms}^{-2}$

[ 2 ]

$a_{0}=\omega^{2} x_{0}$ $\begin{aligned} =9.7^{2} \times 3.4 \times 10^{-2} =3.2 \mathrm{~m} \mathrm{~s}^{-2} \end{aligned}$

(d)

On Fig. 4.2, sketch the variation with x of the acceleration a of the sphere.

Fig. 4.2

[ 3 ]

sketch: straight line through the origin with negative gradient
line with negative gradient passing through ( $+3.4,-a_{0}$ ) and ( $-3.4,+a_{0}$ )
line with ends at $x= \pm 3.4 \mathrm{~cm}$ and $a= \pm a_{0}$

(e)

Describe, without calculation, the interchange between the potential energy and the kinetic energy of the oscillations.

[ 3 ]

[Maximum number: 10]

An object is suspended from a vertical spring as shown in Fig. 3.1.

Fig. 3.1

The object is displaced vertically and then released so that it oscillates, undergoing simple harmonic motion.

Fig. 3.2 shows the variation with displacement x of the energy E of the oscillations.

Fig. 3.2

The kinetic energy, the potential energy and the total energy of the oscillations are each represented by one of the lines P, Q and R .

(a)

State the energy that is represented by each of the lines P, Q and R.

P

Q

R

[ 2 ]

(b)

The object has a mass of 130 g .

Determine the period of the oscillations.
period =

[ 4 ]

$E=1 / 2 m \omega^{2} x_{0}{ }^{2} \quad$ or $\quad E=1 / 2 m v_{0}{ }^{2} \quad$ and $\quad v_{0}=\omega x_{0}$ $\begin{aligned} 6.4 \times 10^{-3}=1 / 2 \times 0.130 \times \omega^{2} \times 0.015^{2} \left(\omega^{2}=438\right) (\omega=20.9) \end{aligned}$ $T=2 \pi / \omega$ $\begin{aligned} =2 \pi / 20.9 =0.30 \mathrm{~s} \end{aligned}$

(c)

State the cause of damping.

[ 1 ]

(d)

A light card is attached to the object. The object is displaced with the same initial amplitude and then released. During each complete oscillation the total energy of the system decreases by 8.0 % of the total energy at the start of that oscillation.

Determine the decrease in total energy, in mJ , of the system by the end of the first 6 complete oscillations.
energy lost = mJ

[ 2 ]

$0.92^{6}$ $\begin{aligned} \text { decrease in energy } =6.4-\left(6.4 \times 0.92^{6}\right) =2.5 \mathrm{~mJ} \end{aligned}$

(e)

State, with a reason, the type of damping that the card introduces into the system.

[ 1 ]

[Maximum number: 8]

A pendulum consists of a bob (small metal sphere) attached to the end of a piece of string. The other end of the string is attached to a fixed point. The bob oscillates with small oscillations about its equilibrium position, as shown in Fig. 4.1.

Fig. 4.1 (not to scale)

The length L of the pendulum, measured from the fixed point to the centre of the bob, is 1.24 m .
The acceleration a of the bob varies with its displacement x from the equilibrium position as shown in Fig. 4.2.

Fig. 4.2

(a)

State how Fig. 4.2 shows that the motion of the pendulum is simple harmonic.

[ 2 ]

(b)

Use Fig. 4.2 to determine the angular frequency $\omega$ of the oscillations.

\omega=\ldots \ldots \ldots \ldots \ldots . \ldots \ldots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ~ r a d ~ s^{-1}

[ 2 ]

$a_{0}=\omega^{2} x_{0}$

or
$a=-\omega^{2} x$

or
$\omega^{2}=- \text { gradient }$ $\begin{aligned} \omega =\sqrt{ }(0.40 / 0.050) =2.8 \mathrm{rad} \mathrm{~s}^{-1} \end{aligned}$

(c)

The angular frequency $\omega$ is related to the length L of the pendulum by

\omega=\sqrt{\frac{k}{L}}

where k is a constant.

Use your answer in (b)(i) to determine k. Give a unit with your answer.

k =unit..............

[ 2 ]

$\begin{aligned} k =\omega^{2} L =2.8^{2} \times 1.24 \end{aligned}$ $=9.7 \mathrm{~m} \mathrm{~s}^{-2}$

(d)

While the pendulum is oscillating, the length of the string is increased in such a way that the total energy of the oscillations remains constant.

Suggest and explain the qualitative effect of this change on the amplitude of the oscillations.

[ 2 ]

(increasing L causes) $\omega$ to decrease

or energy ( $=1 / 2 m \omega^{2} x_{0}{ }^{2}$ ) $=1 / 2 m k x_{0}{ }^{2} / L$ (and L increases)
so amplitude increases

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Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (iii)

Question (b)

Question 2

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Question (c)

Question (d)

Question (e)

Question 3

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Question (d)

Question (e)

Question 4

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Question (d)