Oscillations: Question 6

Worked solution

Step 1: Recall the relationship between angular frequency and frequency

$\omega = 2\pi f$

Substituting $f=200\text{ Hz}$: $\omega = 2\pi \times 200 = 1256.6\text{ rad s}^{-1}\ (5\text{ s.f.})$

Step 2: Recall the maximum acceleration in SHM

The acceleration of a particle in SHM at displacement $x$ is $a=-\omega^2x$. This has its greatest magnitude when $x=x_0$ (the particle is at its extreme displacement, where the restoring force is largest), giving: $a_{max}=\omega^2x_0$

Step 3: Substitute the values

First convert $x_0=1.5\text{ mm}=1.5\times10^{-3}\text{ m}$.

$\omega^2 = (1256.6)^2 = 1.5791\times10^6\text{ rad}^2\text{ s}^{-2}$ $a_{max} = 1.5791\times10^6 \times 1.5\times10^{-3} = 2368.7\text{ m s}^{-2}$

Recompute as a check, using $\omega^2=4\pi^2f^2$ directly: $4\pi^2\times(200)^2=4\pi^2\times40000=1.5791\times10^6\text{ rad}^2\text{ s}^{-2}$, and $1.5791\times10^6\times1.5\times10^{-3}=2368.7\text{ m s}^{-2}$, the same result.

Rounding to 3 significant figures, $a_{max}\approx\boxed{2.37\times10^3}\text{ m s}^{-2}$.

Why the other options are wrong

• B ($4.74\times10^3\text{ m s}^{-2}$): comes from using the peak-to-peak swing $2x_0=3.0\times10^{-3}\text{ m}$ in place of the amplitude $x_0=1.5\times10^{-3}\text{ m}$: $1.5791\times10^6\times3.0\times10^{-3}=4737.4\text{ m s}^{-2}$. The amplitude is the displacement from the centre to one extreme, not the full swing.
• C ($60.0\text{ m s}^{-2}$): comes from treating $\omega=f=200\text{ rad s}^{-1}$ (forgetting the factor of $2\pi$): $(200)^2\times1.5\times10^{-3}=60.0\text{ m s}^{-2}$.
• D ($1.89\text{ m s}^{-2}$): comes from forgetting to square $\omega$, using $a_{max}=\omega x_0=1256.6\times1.5\times10^{-3}=1.885\text{ m s}^{-2}$ instead of $a_{max}=\omega^2x_0$.

Final answer

• The maximum acceleration of the cone is $\boxed{2.37\times10^3}\text{ m s}^{-2}$, option A.