InterviewSolution
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InterviewSolution
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A tunnig fork of frequency 256 Hz produces 4 beats per second with a wire of length 25 cm vibrating in its fundamental mode. The beat frequency decreases when the length is slightly shortened. What could be the minimum length by which the wire be shortened so that it produces no beats with the tuning fork ? |
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Answer» Correct Answer - C Given that `I = 25 cm = 25 xx 10^(-2)m` By shortening the wire the frequency increses `[f = ((1)/(2I)) sqrt((T)/(m))]` As the vibrating wire produces 4 beats with 256 Hz, its frequency must be 252 Hz or 260 Hz. Its frequecy must be 252 Hz, beacuse beat frequecy decreases by shortening the wire. So, `252 = (1)/(2 xx 25 xx 10^(-2)) sqrt((T)/(M)) ... (1)` Let length of the wire be I, after is slightly shortened, `rArr 252 = (1)/(2 xx I_1) sqrt((T)/(M))......(2)` Dividing (1) by (2), we get, `(252)/(256) = (I_1)/(2xx 25 xx 10^(-2))` `rArr I_1 = (252 xx 2 xx 25 xx 10^(-2))/(256)` = 0.24609 m So it should be shorten by (25 - 24.61) = 0.39 cm. |
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