Question

A particle executes simple harmonic motion with an amplitude of 4 cm. At the mean position, the velocity of the particle is 10 cm/s. The distance of the particle from the mean position when its speed becomes 5 cm/s is $\sqrt{\alpha}$ cm, where $\alpha$ = ______

Answer 1

At mean position:

$v_{\text{max}} = A \omega = 10 \Rightarrow \omega = \frac{10}{4} = \frac{5}{2}$

Using the equation $v = \omega \sqrt{A^2 - x^2}$ and $v = 5$:

$5 = \frac{5}{2} \sqrt{4^2 - x^2}$

Solving,

$\sqrt{A^2 - x^2} = 2 \Rightarrow x^2 = A^2 - 4 = 16 - 4 = 12$

Thus, $x = \sqrt{12}$ and $\alpha = 12$.