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If two sound waves, y_1 = 0.3 sin 596 pi [t - x//330] and y_2 =0.5 sin 604 pi[t - x//330] are superposed, what will be the (a) frequency of resultant wave (b) frequency at which the amplitude of resultant waves varies (c ) frequency at which beats aer produced. Find also the ratio of maximum and minimum intensities of beats. |
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Answer» Solution :Comparing the given wave equation with `y =A sin(ometat - kx) = A sin omega [t - (X //v)]` `[as k//omega =I//v]` we FIND that here `A_1 = 0.3 and omega_1 = 2PI f_1 =596 pi` `i.e.,f_1 =298 Hz and A_2=0.5 and omega_2 =2 pi f_2 = 604 pi` `i.e., f_2 = 302 Hz` So (a) The frequency of the resultant wave `f_(av) =(f_1 + f_2)/(2)= ((298+302))/2 = 300 Hz` (b) The frequency at which AMPLITUDE of resultant wave varies: `f_A =(f_1 + f_2)/(2)=((298 - 302))/2 =2Hz` (c ) The frequency at which beats are produced `f_b = 2f_A = f_1 = f_2 = 4Hz` (d) The ratio of maximum to minimum intensities of BEAT `I_(max)/(I_(min)) =((A_1 + A_2)^2)/((A_1 - A_2)^2) =((0.3 + 0.5)^2)/((0.3 - 0.5)^2) =64/4 =16` |
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