InterviewSolution
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InterviewSolution
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Three sound sources `A`, `B` and `C` have frequencies `400`, `401` and `402 H_(Z)`, respectively. Cacluated the number of beats noted per second. |
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Answer» Correct Answer - A::B::C Let as make the following table. Beat time period for `A` and `B` is `1` s. It implies that `A` and `B` are in phase at time `t = 0`, they are again in phase after `1` s. Same is the case with `B` and `C` . But beat time period for `A` and `C` is `0.5`s. Therefore , beat time period for all together `A, B` and `C` will be `1` s. Because if , at `t = 0, A, B` and `C` all are in phase then after `1` s. (`A` and `B`) and (`B` and `C`) will again be in phase for the first time while (`A` and `C`) will be in phase for the second time. Or we can that all `A, B` and `C` are again in phase after `1` s. `:.` Beat time period , `T_(b) = 1 s` or Beat frequency , `f_(b) = (1)/(T_(b)) = H_(Z)` Suppose at time `t`, the equations of waves are `y_(1) = A_(1) sin 2 pif_(A)t` `(omega = 2pi f)` `y_(2) = A_(2) sin 2pi f_(B) t` and `y_(3) = A_(3) sin 2pi f_(C) t` If they are beat time at some given instant of time `t`, then `2 pif_(A)t = 2 pi f_(B)t = 2pi f_(C)t` ...(i) Let `T_(b)` be the beat time period, i.e. after time `T_(b)` they all are again in phase. As `f_(C) gt f_(B) gt f_(A)` , so `2pi f_(C) (t + T_(b)) = 2 pi f_(A) (t + T_(b)) + 2 mpi` ...(ii) and `2pi f_(B) (t + T_(b)) = 2 pi f_(A) (t + T_(b)) + 2 npi` ....(iii) Here, `m` and `n (lt m)` are positive integers. From Eqs. (i) and (ii) `(f_(C) - f_(A)) T_(b) = m` ...(iv) Similary, From Eqs. (i) and (ii) `(f_(B) - f_(A)) T_(b) = n` ...(v) Dividing Eq. (iv)by Eq. (v), `(m)/(n) = (f_(C) - f_(A))/(f_(B) -f_(A))= (402 - 400)/(401 - 400)= (2)/(1)` Thus, letting `m = 2` and `n = 1` `T_(b)= (m)/(f_(C) - f_(A))` [from Eq. (iv)] = `(2)/(2) = 1 H_(Z)` `:.` Beat frequency , `f_(b) = (1)/(T_(b)) = 1 H_(Z)` |
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