# IBDP Physics 9.1 – Simple harmonic motion : IB Style Question Bank HL Paper 1

### Question

Which is correct for the tangential acceleration of a simple pendulum at small amplitudes?

A It is inversely proportional to displacement.

B It is proportional to displacement.

C It is opposite to displacement.

D It is proportional and opposite to displacement.

Answer/Explanation

Ans: D

A simple pendulum consists of a point mass suspended by a weightless inextensible cord from a rigid support.
Let a bob of mass m is displaced from its, equilibrium position and released, then it oscillates in a vertical plane under gravity. Let θ be the angular displacement at any time t, then corresponding linear displacement along the arc is
x = l θ.
It is clear from the diagram that mg sinθ, is the restoring force acting on m tending to return it to mean position. So from Newton’s second law
…(i)
where negative sign indicates that restoring force mg sin θ (= F) is opposite to displacement θ. If θ is very small, then
sin θ ≈ θ, so from equation (i)
…(ii)

### Question

A simple pendulum bob oscillates as shown.

At which position is the resultant force on the pendulum bob zero?

A. At position A

B. At position B

C. At position C

D. Resultant force is never zero during the oscillation

Answer/Explanation

## Markscheme

D

At the equilibrium position of an oscillating pendulum, its acceleration is momentarily zero as the velocity reaches its maximum value. But this is circular motion, meaning that it must always have a resultant force towards the centre of the circle.

### Question

A mass at the end of a vertical spring and a simple pendulum perform oscillations on Earth that are simple harmonic with time period T. Both the pendulum and the mass-spring system are taken to the Moon. The acceleration of free fall on the Moon is smaller than that on Earth. What is correct about the time periods of the pendulum and the mass-spring system on the Moon?

Answer/Explanation

### Markscheme

B

Simple pendulum
This is the equation of S.H.M. of the bob with time period
Spring mass system
Here we can see , for Simple pendulum Time period is inversely proportional to g
or
$$T \propto \frac{1}{\sqrt{g}}$$
Hence g decreases means T increases
But in Spring mass system
T is independent of g hence , here Time period T remain constant.

### Question

A spring loaded with mass m oscillates with simple harmonic motion. The amplitude of the motion is A and the spring has total energy E. What is the total energy of the spring when the mass is increased to 3m and the amplitude is increased to 2A?

A. 2E

B. 4E

C. 12E

D. 18E

Answer/Explanation

### Markscheme

B

Total mechanical energy
E = K.E. + P.E.
Hence Total Energy depends on $$k$$ of spring and Amplitude $$A$$ of Oscillation
Given  $$A’ = 2A$$
$$\Rightarrow E’ = \frac{1}{2}kA’^2 = \frac{1}{2}k 4 A^2$$
$$=4E$$

### Question

A mass oscillates with simple harmonic motion (SHM) of amplitude xo. Its total energy is 16 J.

What is the kinetic energy of the mass when its displacement is $$\frac{{{x_0}}}{2}$$?

A. 4 J

B. 8 J

C. 12 J

D. 16 J

Answer/Explanation

### Markscheme

C

Total Energy $$= \frac{1}{2}kx_0^2$$
Kinetic Energy $$= \frac{1}{2}k(x_0^2-x^2)$$
Given $$x= \frac{x_0}{2}$$
Hence
K.E when $$x = \frac{x_0}{2} =$$
$$\frac{1}{2}k(x_0^2-x^2) =\frac{1}{2}k(x_0^2-(\frac{x_0}{2})^2)$$
$$=\frac{3}{4}(\frac{1}{2}kx_0^2) = \frac{3}{4} E$$
Also $$E= 16 \;J$$
Hence
$$K.E = \frac{3}{4}\times 16 =12 \;J$$

### Question

A pendulum oscillating near the surface of the Earth swings with a time period T. What is the time period of the same pendulum near the surface of the planet Mercury where the gravitational field strength is 0.4g?

A.  0.4T

B.  0.6T

C.  1.6T

D.  2.5T

Answer/Explanation

### Markscheme

C

This is the equation of S.H.M. of the bob with time period
Hence
$$T_{earth}=2\pi\sqrt{\frac{l}{g_{earth}}}$$
$$T_{Merc}=2\pi\sqrt{\frac{l}{g_{Merc}}}$$
or
$$\frac{T_{Merc}}{T_{earth}}=\sqrt{\frac{g_{earth}}{g_{Merc}}}$$
$$=\sqrt{\frac{g}{0.4g}}=\sqrt{2.5}$$
$$T_{Merc}\approx 1.6T$$

### Question

A particle executes simple harmonic motion (SHM) with period T.

Which sketch graph correctly shows how the total energy E of the particle varies with time t from t = 0 to $$t = \frac{T}{2}$$?

Answer/Explanation

## Markscheme

D

Total Energy is constant.

### Question

A particle of mass $$m$$ oscillates with simple harmonic motion (SHM) of angular frequency $$\omega$$. The amplitude of the SHM is $$A$$. What is the kinetic energy of the particle when it is half way between the equilibrium position and one extreme of the motion?

A.     $$\frac{{m{A^2}{\omega ^2}}}{4}$$

B.     $$\frac{{3m{A^2}{\omega ^2}}}{8}$$

C.     $$\frac{{9m{A^2}{\omega ^2}}}{{32}}$$

D.     $$\frac{{15m{A^2}{\omega ^2}}}{{32}}$$

Answer/Explanation

### Markscheme

B

Kinetic energy : A particle in S.H.M. possesses kinetic energy by virtue of its motion.
= $$K(t)=\frac{1}{2}mv^2 =\frac{1}{2}mw^2x_m^2sin^2(wt+\phi )$$
Given in the question  $$y=\frac{A}{2}$$
$$\therefore K.E = \frac{1}{2}mw^2(A^2-(\frac{A}{2})^2 )$$
$$=\frac{3}{8}mw^2A^2$$