A) \[\frac{1}{6}s\]
B) \[\frac{1}{4}s\]
C) \[\frac{1}{3}s\]
D) \[\frac{1}{12}s\]
Correct Answer: A
Solution :
The kinetic energy of a body undergoing SHM is given by \[KE=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}{{\cos }^{2}}\omega t\]and\[K{{E}_{\max }}=\frac{m{{\omega }^{2}}{{A}^{2}}}{2}\] (as maximum value of\[\cos \omega t=1\]and here symbols represent standard quantities) From given information, \[KE=(K{{E}_{\max }})\times \frac{75}{100}\] \[\Rightarrow \]\[\frac{m{{\omega }^{2}}{{A}^{2}}}{2}{{\cos }^{2}}\omega t=\frac{m{{\omega }^{2}}{{A}^{2}}}{2}\times \frac{3}{4}\] \[\Rightarrow \]\[\cos \omega t=\pm \frac{\sqrt{3}}{2}\]\[\Rightarrow \]\[\omega t=\frac{\pi }{6}\] \[\Rightarrow \]\[\frac{2\pi }{T}\times t=\frac{\pi }{6}\]\[\Rightarrow \]\[t=\frac{T}{12}=\frac{1}{6}s\]
You need to login to perform this action.
You will be redirected in
3 sec
