

InterviewSolution
Saved Bookmarks
1. |
A particle performing S.H.M. about their mean position with the equation of velocity is given by `4v^(2)=25-x^(2)`, then the period of motion isA. `2pi`B. `pi`C. `3pi`D. `4pi` |
Answer» Correct Answer - B `4v^(2)=25-x^(2)` `v^(2)=(1)/(4)[25-x^(2)]` `v=pm(1)/(2)sqrt(25-x^(2))` Compare it with standard equation of velocity, `v=omegasqrt(A^(2)-x^(2))` `omega=(1)/(2) therefore T=(2pi)/(omega)=(pi)/(1//2)` `therefore T=4pis` |
|