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)

$\frac{ \sqrt 3 A }{ 2}$

Answer 2

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