Oscillations: Question 1

Worked solution

Step 1: Recall the relationship between angular frequency and frequency

$\omega = 2\pi f$

Substituting $f=3.0\text{ Hz}$: $\omega = 2\pi \times 3.0 = 18.8\text{ rad s}^{-1}\ (3\text{ s.f.})$

Step 2: Recall the maximum speed in SHM

The speed of a particle in SHM at displacement $x$ is $v=\pm\omega\sqrt{x_0^2-x^2}$. This is greatest when $x=0$ (the particle passes through the centre of the oscillation), giving: $v_{max}=\omega x_0$

Step 3: Substitute the values

$v_{max} = 18.85 \times 0.045 = 0.848\text{ m s}^{-1}$

Recompute as a check, multiplying in a different order: $x_0 \times 2\pi f = 0.045\times2\times\pi\times3.0 = 0.045\times18.850=0.848\text{ m s}^{-1}$. Both routes agree.

Rounding to 2 significant figures, $v_{max}\approx\boxed{0.85}\text{ m s}^{-1}$.

Why the other options are wrong

• B ($1.7\text{ m s}^{-1}$): comes from using the peak-to-peak swing $2x_0=0.090\text{ m}$ in place of the amplitude $x_0=0.045\text{ m}$: $18.85\times0.090=1.70\text{ m s}^{-1}$. The amplitude is the displacement from the centre to one extreme, not the full swing.
• C ($0.14\text{ m s}^{-1}$): comes from treating $\omega=f=3.0\text{ rad s}^{-1}$ (forgetting the factor of $2\pi$): $3.0\times0.045=0.135\text{ m s}^{-1}$.
• D ($0.094\text{ m s}^{-1}$): comes from inverting the relationship, using $\omega=2\pi/f=2.094\text{ rad s}^{-1}$ instead of $\omega=2\pi f$: $2.094\times0.045=0.094\text{ m s}^{-1}$.

Final answer

• The maximum speed of the ball is $\boxed{0.85}\text{ m s}^{-1}$, option A.