Question
A simple harmonic oscillator has an amplitude $$a$$ and time period $$T.$$ The time required by it to travel from $$x = a$$ to $$x = \frac{a}{2}$$ is
A.
$$\frac{T}{6}$$
B.
$$\frac{T}{4}$$
C.
$$\frac{T}{3}$$
D.
$$\frac{T}{2}$$
Answer :
$$\frac{T}{6}$$
Solution :
Equation of $$SHM$$ is
$$x = a\sin \omega t\,\,{\text{or}}\,\,x = a\sin \left( {\frac{{2\pi }}{T}} \right)t$$
when $$x = a,$$ then
$$\eqalign{
& a = a\sin \left( {\frac{{2\pi }}{T}} \right)t \cr
& {\text{or}}\,\,\sin \left( {\frac{{2\pi }}{T}} \right)t = 1 \cr
& {\text{or}}\,\,\sin \left( {\frac{{2\pi }}{T}} \right)t = \sin \frac{\pi }{2} \cr
& \Rightarrow t = \frac{T}{4} \cr
& {\text{when}}\,x = \frac{a}{2},\,{\text{then}} \cr
& \frac{a}{2} = a\sin \left( {\frac{{2\pi }}{T} \cdot t} \right) \cr
& {\text{or}}\,\,\sin \left( {\frac{{2\pi }}{T}t} \right) = \sin \frac{\pi }{6}\,\,{\text{or}}\,t = \frac{T}{{12}} \cr} $$
Hence, time taken to travel from
$$x = a\,\,{\text{to}}\,\,x = \frac{a}{2} = \frac{T}{4} - \frac{T}{{12}} = \frac{T}{6}$$