Q 11-13-036JEE MainMedium
A particle executes SHM of amplitude $4$ cm. Its speed at the mean position is $12$ cm/s. Find its speed, in cm/s, when it is $2\sqrt{3}$ cm from the mean position.
Numerical value type. Enter your answer.
Answer: 6
Concept. $v_{\max} = A\omega$ gives $\omega$; then $v = \omega\sqrt{A^2 - x^2}$.
Step 1. Angular frequency. $\omega = \dfrac{12}{4} = 3$ rad/s.
Step 2. Speed. $v = 3\sqrt{16 - 12} = 6$ cm/s.
Answer. $6$.
Solution by Sreeraj P, M.Sc Physics