InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
The ends of a stretched wire of length `L` are fixed at `x = 0 and x = L`. In one experiment, the displacement of the wire is `y_(1) = A sin(pi//L) sin omegat` and energy is `E_(1)` and in another experiment its displacement is `y_(2) = A sin (2pix//L ) sin 2omegat` and energy is `E_(2)`. ThenA. (a) `E_(2) = E_(1)`B. (b) `E_(2) = 2E_(1)`C. (c) `E_(2) = 4E_(1)`D. (d) `E_(2) = 16E_(1)` |
|
Answer» Correct Answer - C ( c ) `E prop A^(2)v^(2)` where `A = amplitude and v = frequency`. Also `omega = 2piv rArr omega prop v` In case 1 : `Amplitude = A and v_(1) = v` In case 2 : `Amplitude = A and v_(2) = 2v` :. `(E_(2))/(E_(1)) = (A^(2)v_(2)^(2))/(A^(2)v_(1)^(2)) = 4` rArr `E_(2) = 4E_(1)` |
|