# A-level Physics (Advancing Physics)/Simple Harmonic Motion/Worked Solutions

1. A 10N weight extends a spring by 5 cm. Another 10N weight is added, and the spring extends another 5 cm. What is the spring constant of the spring?

${\displaystyle \Delta F=k\Delta x}$

${\displaystyle 10=0.05k}$

So ${\displaystyle k={\frac {10}{0.05}}=200Nm^{-1}}$

2. The spring is taken into outer space, and is stretched 10 cm with the two weights attached. What is the time period of its oscillation?

First calculate the mass of the two weights:

${\displaystyle 20=9.81m}$

m = 2.04 kg

${\displaystyle T=2\pi {\sqrt {\frac {m}{k}}}=2\pi {\sqrt {\frac {2.04}{200}}}=0.634{\mbox{ s}}}$

3. What force is acting on the spring after 1 second? In what direction?

We are starting the oscillation at t=0 with a displacement. This displacement is the amplitude of the oscillation, and we need f(ωt) to be positive at t=0. So, we use an equation for the displacement with a cosine in it. We have already derived the acceleration in this case:

${\displaystyle a=-\omega ^{2}\cos {\omega t}}$

${\displaystyle \omega ={\frac {2\pi }{T}}={\frac {2\pi }{6.34}}=0.99{\mbox{ rad s}}^{-1}}$

So:

${\displaystyle a=-0.99^{2}\cos {0.99\times t}=-0.538{\mbox{ ms}}^{-2}}$

The minus sign means that the acceleration is in the opposite direction to the initial displacement.

4. A pendulum oscillates with a frequency of 0.5 Hz. What is the length of the pendulum?

${\displaystyle \omega =2\pi f}$

${\displaystyle {\sqrt {\frac {g}{l}}}=2\times \pi \times 0.5}$

${\displaystyle {\sqrt {\frac {9.81}{l}}}=\pi }$

${\displaystyle {\frac {9.81}{l}}=\pi ^{2}}$

${\displaystyle l={\frac {9.81}{\pi ^{2}}}=0.994{\mbox{ m}}}$

5. The following graph shows the displacement of a simple harmonic oscillator. Draw graphs of its velocity, momentum, acceleration and the force acting on it.

x is a sine wave, so:

${\displaystyle x=A\sin {\omega t}}$

${\displaystyle v={\frac {dx}{dt}}=A\omega \cos {\omega t}}$

${\displaystyle a={\frac {dv}{dt}}={\frac {d^{2}x}{dt^{2}}}=-A\omega ^{2}\sin {\omega t}}$

${\displaystyle p=mv=mA\omega \cos {\omega t}}$

${\displaystyle F=ma=-mA\omega ^{2}\sin {\omega t}}$

Since you haven't been given any details, the amplitudes of the waves don't matter. The phase differences, however, do.

6. A pendulum can only be modelled as a simple harmonic oscillator if the angle over which it oscillates is small. Why is this?

Simple harmonic oscillators work because the force acts in the opposite direction to the displacement. As the pendulum moves away from the area immediately below the peg it is hanging on, the force no longer acts in the opposite direction to the displacement.