InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The transverse displacement of a string clamped at its both ends is given by `y(x, t) = 0.06 sin ((2pi)/3 x) cos(l20pit)` where x and y are in m and t in s. The length of the string is 1.5 m and its mass is `3 xx 10^(-2)` kg. The tension in the string isA. 324 NB. 648 NC. 832 ND. 972 N |
|
Answer» Correct Answer - B The given equation is `y(x,t)=0.06 sin((2pi)/3x)cos(120pit)` Compare it with y(x, t) = 2a sin kx cos `omega`t we get, k`=(2pi)/3or(2pi)/lamda=(2pi)/3orlamda=3 m` and `omega = 120pi: or 2piupsilon = 120pi or upsilon =- 60 Hz = 60 s^(-1)` Velocity of wave, `v = upsilonlamda = (60 s^(- 1)) (3 m) = 180 m s^(-1)` Mass per unit length of the string, `mu=(3xx10^(-2)kg)/(1.5m)=2xx10^(-2)kgm^(-1)` Velocity of transverse wave in the string, v `v=sqrt(T/mu)orv^(2)=T/muorT=v^(2)mu` `T=(180ms^(-1))^(2)(2xx10^(-2)kgm^(-1))=648 N` |
|