Question

A particle executes S.H.M. with amplitude 'a', and time period 'T'. The displacement of the particle when its speed is half of maximum speed is \(\frac{\sqrt{x}\,a}{2}\). The value of \(x\) is __________.

Hint:

Use \(v = \omega\sqrt{a^2 - y^2}\) with \(v = \omega a/2\).

Answer 1

Correct Answer: 3

In S.H.M. the speed at displacement \(y\) is \(v = \omega\sqrt{a^2 - y^2}\), and the maximum speed is \(v_{max} = \omega a\).
Given \(v = \frac{v_{max}}{2} = \frac{\omega a}{2}\).
\(\frac{\omega a}{2} = \omega\sqrt{a^2 - y^2}\) so \(\frac{a^2}{4} = a^2 - y^2\), giving \(y^2 = \frac{3a^2}{4}\).
\(y = \frac{\sqrt{3}\,a}{2}\). Comparing with \(\frac{\sqrt{x}\,a}{2}\): \(x = 3\).
Answer: 3