Question

A particle starts with S.H.M. from the mean position as shown in the figure. Its amplitude is $A$ and its time period is $T$. At one time, its speed is half that of the maximum speed. What is this displacement?

• A. $ \frac{ 2 A}{ \sqrt 3} $
• B. $ \frac{ 3 A}{ \sqrt 2} $
• C. $ \frac{ \sqrt 2 A }{ 3} $
• D. $ \frac{ \sqrt 3 A }{ 2} $

Answer 1

Answer: (D)

Maximum velocity, $ v_{ \max} = = A \omega$
According to question, $ \frac{ v_{ \max}}{ 2} = \frac{ A \omega}{ 2} = \omega \sqrt{ A^2 - y^2 } $
$ \frac{ A^2 }{ 4} = A^2 - y^2 $
$\Rightarrow y^2 = A^2 - \frac{ A^2}{4} $
$\Rightarrow y = \frac{ \sqrt 3 \, A}{2}$.