Saved Bookmarks
| 1. |
A particle performs SHM with amplitude 25 cm and period 3s the minimu time required for it to move between two points 12.5 cm on either side of the mean position isA. 0.6 sB. 0.5 sC. 0.4 sD. 0.2 s |
|
Answer» Correct Answer - d Here amplitude of particle A= 25 cm and time period T= 3s if the paritcle at t=0 s is at mean position its displacement equation will be `x=A sin omega t` if it takes time `t_(1)` to move a distance x= 12.5 cm to one side of its mena position then `12.5 =25 sin (2pit_(1))/(3)` `1/2=-sin(2pit_(1))/(3) ro sin (pi)/(6)=sin (2pit_(1))/(3)` `therefore (pi)/(6)=(2pit_(1))/(3)rarr t_(1)=1/4` s The same will be the time to move 12.5 cm to the other side of its mean position therefore total time `t=t_(1)+t_(2)=1/4+1/4=1/2=0.5 `s |
|