Question

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is

[Haryana CEE 1996; CBSE PMT 1996; MH CET 2002]

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$v_{max}=\omega A\Rightarrow v=\frac{\omega A}{2}=\omega \sqrt{A^2-y^2}$

$\Rightarrow A^2-y^2=\frac{A^2}{4}\Rightarrow y^2=\frac{3A^2}{4}\Rightarrow y=\frac{\sqrt{3}A}{2}$