The frequency of a tuning fork A is 5% greater than that of a standard fork K. The frequency of another fork B is 3% less than that of K. When A and B are vibrated simultaneously 4 beats per second are heard. Find the frequencies of A and B.

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124 Hz, 120 Hz

63.5 Hz, 79.5 Hz

10.5 Hz, 101 Hz

52.5 Hz, 48.5 Hz

answer is A.

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Detailed Solution

$\large {n_A}\, = \,\frac{{105}}{{100}}{n_k}\,\,;\,\,{n_B}\, = \,\frac{{97}}{{100}}{n_k}\,;\,{n_A} - {n_B}\, = \,4$

$\large \frac{{105}}{{100}}{n_k} - \frac{{97}}{{100}}{n_k}\, = \,4;\,\frac{{8{n_k}}}{{100}}\, = \,4$

$\large {n_k}\, = \,50Hz\,\,\,\,\therefore {n_A}\, = \,\frac{{105}}{{100}} \times 50\, = \,52.5Hz$

$\large \therefore \,{n_B}\, = \,\frac{{97}}{{100}} \times 50\, = \,48.5Hz$

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