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

Correct Answer: 12

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 \( \alpha = 12 \).