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There are three source of sound of equal intensities with frequencies 400, 401 and 402 Hz. The no. of beats per second is : |
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Answer» 0 ` thereforey_(1) = r sin 2pi 400 t , y_(2)= r sin 2 pi401 t ` `y_(3) = r sin 2pi 402 t.` Now y = `y_(1) + y_(3) + y_(2)` y = r sin `2pi` 400t + r sin `2pi` 402 t + r sin `2pi` 401 t. y = r sin `2 pi ((402 - 400)/(2)) t cos 2pi ((402 - 400)/(2) ) + r sin 2 pi 401 t`. `y = 2r sin 2pi 401 t cos 2pi t + r sin 2 pi 401 t`. y = `[2 cos 2 PIT + 1] r sin 2 pi 401 t`. Now Amplitude of resultant wave is ` A = (2 cos 2pi t + 1) r. ` For A to be maximum , cos `2 pi t = + 1 ` `rArr "" 2 pi t = 2 n pi, n = 0,1,2, t = n ` Amplitude will be maximum at time t = 0 , 1,2,3 ... sec. Time interval between successive maximum = 1 sec. `therefore` BEAT FREQUENCY = 1 Hz. hence the CORRECT choice is (b) . |
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