question_answer

A particle executes SHM               of amplitude A. If \[{{T}_{1}}\] and \[{{T}_{2}}\] are the times taken by the particle to traverse from 0 to\[\frac{A}{2}\]and from \[\frac{A}{2}\]to A respectively, then\[\frac{{{T}_{1}}}{{{T}_{2}}}\] will be equal to

A) 1             

B)        \[\frac{1}{2}\]                   

C) \[\frac{1}{4}\]                                   

D) \[2\]

Correct Answer: B

Solution :

The displacement equation of the SHM is \[y=a\sin \omega t\] where a is amplitude and co is angular velocity when time taken is\[{{T}_{1}}\] from 0 to \[\frac{A}{2},\] then \[\frac{A}{2}=A\sin \frac{2\pi {{T}_{1}}}{T}\] \[\Rightarrow \]                               \[\sin \frac{2\pi {{T}_{1}}}{T}=\frac{1}{2}\] \[\Rightarrow \]                               \[\frac{2\pi {{T}_{1}}}{T}=\frac{\pi }{6}\]                \[\Rightarrow \]                               \[{{T}_{1}}=\frac{T}{12}\] Time taken by particle to traverse one-fourth cycle, ie, from \[\frac{A}{2}\] to A. \[\therefore \]  \[\frac{T}{4}={{T}_{1}}+{{T}_{2}}\] \[\Rightarrow \]               \[{{T}_{2}}=\frac{T}{4}-{{T}_{1}}=\frac{T}{4}-\frac{T}{12}\]                 \[=\frac{3T-T}{12}=\frac{T}{6}\] Hence,                  \[\frac{{{T}_{1}}}{{{T}_{2}}}=\frac{T/12}{T/6}=\frac{1}{2}\]