Oscillations: Question 7

Worked solution

Part (a): Choosing the displacement equation

At $t=0$ the prong tip is stated to pass through the centre of the oscillation ($x=0$) with maximum speed. The equation $x=x_0\sin\omega t$ satisfies this: at $t=0$, $\sin(0)=0$ so $x=0$, and the velocity $v=x_0\omega\cos\omega t$ is $x_0\omega\cos(0)=x_0\omega$, its maximum value. So: $x=x_0\sin\omega t$

By contrast, $x=x_0\cos\omega t$ would give $x=x_0$ (maximum displacement) and $v=0$ at $t=0$. The opposite initial condition, and not what is described here.

Part (b): Angular frequency

$\omega=2\pi f=2\pi\times256=1608.5\text{ rad s}^{-1}\ (5\text{ s.f.})$

Recompute as a check, multiplying in a different order: $256\times2\times\pi=256\times6.2832=1608.5\text{ rad s}^{-1}$, the same value.

So $\omega=1610\text{ rad s}^{-1}$ (3 s.f.). The unrounded value $1608.5\text{ rad s}^{-1}$ is carried forward for accuracy.

Part (c): Displacement and velocity at $t=2.0\times10^{-4}\text{ s}$

First convert $x_0=0.80\text{ mm}=8.0\times10^{-4}\text{ m}$.

Find the phase angle: $\omega t=1608.5\times2.0\times10^{-4}=0.3217\text{ rad}$ (working in radians).

Displacement: $x=x_0\sin(\omega t)=8.0\times10^{-4}\times\sin(0.3217)=8.0\times10^{-4}\times0.3162=2.529\times10^{-4}\text{ m}$

Velocity: $v=x_0\omega\cos(\omega t)=8.0\times10^{-4}\times1608.5\times\cos(0.3217)=1.2868\times0.9487=1.221\text{ m s}^{-1}$

Recompute the velocity as a check, using $v=\pm\omega\sqrt{x_0^2-x^2}$ instead: $x_0^2-x^2=(8.0\times10^{-4})^2-(2.529\times10^{-4})^2=6.40\times10^{-7}-6.40\times10^{-8}=5.76\times10^{-7}$, so $\sqrt{5.76\times10^{-7}}=7.589\times10^{-4}\text{ m}$, giving $v=1608.5\times7.589\times10^{-4}=1.221\text{ m s}^{-1}$, the same value, confirming the result.

So $x=\boxed{2.53\times10^{-4}}\text{ m}$ and $v=\boxed{1.22}\text{ m s}^{-1}$ (3 s.f.).

Part (d): Acceleration at this instant

$a=-\omega^2x=-(1608.5)^2\times2.529\times10^{-4}=-2.5873\times10^6\times2.529\times10^{-4}=-654.4\text{ m s}^{-2}$

Recompute as a check, using $a=-a_{max}\sin(\omega t)$ with $a_{max}=\omega^2x_0=2.5873\times10^6\times8.0\times10^{-4}=2069.8\text{ m s}^{-2}$: $a=-2069.8\times\sin(0.3217)=-2069.8\times0.3162=-654.5\text{ m s}^{-2}$, the same value (small rounding only), confirming the result.

The magnitude of the acceleration is $\boxed{654}\text{ m s}^{-2}$ (3 s.f.). Since $x$ is positive (the prong is displaced to the same side as the direction taken as positive) and $a=-\omega^2x$ is negative, the acceleration acts back toward the equilibrium (centre) position, i.e. in the opposite direction to the displacement, as required for any restoring force in SHM.

Final answers

• (a) $x=x_0\sin\omega t$; correct because the prong starts at the centre with maximum speed
• (b) $\omega=\boxed{1610}\text{ rad s}^{-1}$
• (c) $x=\boxed{2.53\times10^{-4}}\text{ m}$, $v=\boxed{1.22}\text{ m s}^{-1}$
• (d) $a=\boxed{654}\text{ m s}^{-2}$, directed back toward the equilibrium position