InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Three plane sources of sound of frequency n_(1) = 400 Hz, 401 Hz and n_(3) = 402 Hzof equal amplitude a each, are sounded together. A detector receives waves from all the three sources simultaneously.It can detect signals of amplitude gt A. Calculate (a) period ofone complete cycle of intensity received by the detector and (b) time for which the detector remains idle in each cycle of intensity. |
|
Answer» Solution :We know that `y = A sin omega t = A sin (2 PI n t)` The displacements of the medium particles CAUSED by these waves are given as `y_(1) = A sin (800 pi t)` `y_(2) = A sin(802pi t)` `y_(3) = A sin (804 pi t)` The resultant displacement of medium particle at time t is given by `y = y_(1) + y_(2) + y_(3)` = `A[sin (800 pi t) + sin(802 pi t ) + sin (804 pi t)]` `A[sin (800 pi t) + sin(802 pi t ) + sin (804 pi t)]` `A [ 1 + COS 2 pi t ] sin 802 pi t` The resultant is also a plane wave. Let its amplitude be R. The, `y = R sin omega t` Here `R = A ( 1 + cos 2 pi t)` Equation (6.115) shows that the resultant amplitude (or resultant intensity, I infty`R^(2))` varies with time. The resultant amplitude is maxiumum, when ` `2PI t = 0, 2 pi, 4 pi`, ... `t = 0, 1, 2, 3 ....` Hence period of one complete cycle of intensity is one second. (b ) Given that signal is detected when A ge a Thus `cos 2 pi t ge 0` Thus cos `2pi t` should lie either in first quadrant or in fourth quadrant I,e.. Between either 0 to `pi/2 or 3pi/2 and 2pi`. So, during first cycle of intensity, the signal is detected when `0 le t le (1)/(4)` and `(3)/(4) le t le 1` This shows that detector remains ideal from `t = (1)/(4)s` to `t = (3)/(4)s` Therefore, in each cycle of intensity, the detector remains ideal for`(3)/(4) - (1)/(4) = (2)/(4)` = 0.5 s. |
|