Question
A particle starts with $$S.H.M.$$ from the mean position as shown in figure below. Its amplitude is $$A$$ and its time period is $$T.$$ At one time, its speed is half that of the maximum speed. What is the displacement at that time ?
A.
$$\frac{{\sqrt 2 A}}{3}$$
B.
$$\frac{{\sqrt 3 A}}{2}$$
C.
$$\frac{{2A}}{{\sqrt 3 }}$$
D.
$$\frac{{3A}}{{\sqrt 2 }}$$
Answer :
$$\frac{{\sqrt 3 A}}{2}$$
Solution :
$$\eqalign{
& v = \omega {\left[ {{A^2} - {x^2}} \right]^{\frac{1}{2}}} \Rightarrow x = {\left[ {{A^2} - \frac{{{v^2}}}{{{\omega ^2}}}} \right]^{\frac{1}{2}}} \cr
& {\text{Given}}\,{\text{that}}\,\,v = \frac{{{v_{\max }}}}{2} = \frac{{A\omega }}{2}. \cr
& {\text{so,}}\,\,{\left[ {{A^2} - \frac{{{A^2}{\omega ^2}}}{{4{\omega ^2}}}} \right]^{\frac{1}{2}}} = \frac{{\sqrt 3 }}{2}.A \cr} $$