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Two loudspeakers `L_1` and `L_2` driven by a common oscillator and amplifier, are arranged as shown. The frequency of the oscillator is gradually increased from zero and the detector at D records a series of maxima and minima. If the speed of sound is `330ms^-1` then the frequency at which the first maximum is observed isA. `165 Hz`B. `330 Hz`C. `496 Hz`D. `660 Hz` |
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Answer» Correct Answer - B `Delta x = L_(2)D - L_(1)D = sqrt((99)^(2) + (40)^(2)) - 40` `= 41 - 40 = 1 m` For maximum intensity `Delta x = n lambda = n(v)/(f)` `f = (n v)/(Delta x) = (n xx 330)/(1) = 330 n` `n = 1, f = 330 Hz` |
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