Given data

• Amplitude: $A = 4\ \text{cm}$
• Speed at mean position: $v_{\max} = 10\ \text{cm/s}$
• Required speed: $v = 5\ \text{cm/s}$

Use SHM speed relation

For simple harmonic motion, $v = \omega \sqrt{A^2 - x^2}$ where $x$ is the displacement from mean position.

Also, at mean position $x=0$, so speed is maximum: $v_{\max} = \omega A$

Find angular frequency $\omega$ $\omega A = 10$ $\omega \cdot 4 = 10$ $\omega = \frac{10}{4} = 2.5\ \text{s}^{-1}$

Use given speed $v=5\ \text{cm/s}$ $5 = 2.5\sqrt{4^2 - x^2}$ $5 = 2.5\sqrt{16 - x^2}$ $\sqrt{16 - x^2} = 2$ $16 - x^2 = 4$ $x^2 = 12$ $x = \sqrt{12}\ \text{cm}$

Compare with $\sqrt{\alpha}$ Since the distance is given as $\sqrt{\alpha}\ \text{cm}$, $\alpha = 12$

Comparison with stored answer Stored correct answer = $12$

This matches our derived answer.