Question
A particle starts simple harmonic motion from the mean position. Its amplitude is a and time period is $$T.$$ What is its displacement when its speed is half of its maximum speed ?
A.
$$\frac{{\sqrt 2 }}{3}a$$
B.
$$\frac{{\sqrt 3 }}{2}a$$
C.
$$\frac{2}{{\sqrt 3 }}a$$
D.
$$\frac{a}{{\sqrt 2 }}$$
Answer :
$$\frac{{\sqrt 3 }}{2}a$$
Solution :
Velocity of the particle executing $$SHM$$ at any instant is defined as the time rate of change of its displacement at that instant.
Let the displacement of the particle at an instant $$t$$ is given by
$$\eqalign{
& x = a\sin \omega t \cr
& \therefore {\text{velocity}}\,v = \frac{{dx}}{{dt}} = \frac{{d\left( {a\sin \omega t} \right)}}{{dt}} \cr
& = a\omega \cos \omega t = a\omega \sqrt {\left( {1 - {{\sin }^2}\omega t} \right)} \cr
& = a\omega \sqrt {\left( {1 - \frac{{{x^2}}}{{{a^2}}}} \right)} \cr
& = \omega \sqrt {\left( {{a^2} - {x^2}} \right)} \cr} $$
At mean position, $$x = 0$$
$$\therefore {v_{\max }} = \omega a$$
According to problem, $$v = \frac{{{v_{\max }}}}{2} = \frac{{a\omega }}{2}$$
$$\eqalign{
& {\text{But}}\,v = \omega \sqrt {{a^2} - {x^2}} \cr
& \therefore \frac{{a\omega }}{2} = \omega \sqrt {{a^2} - {x^2}} \,\,{\text{or}}\,\,x = \frac{{\sqrt 3 }}{2}a \cr} $$