A) 506 Hz
B) 512 Hz
C) 518 Hz
D) 524 Hz
Correct Answer: A
Solution :
Let original frequency of sonometer wire is f Hz, then f = 512 ± 6 = 518 or 506 Hz. When tension in the string is increased, frequency of string increases, while beat frequency is decreasing. \[512=f+{{f}_{b}}\]or \[f-{{f}_{b}}\]where \[{{f}_{b}}\] is beat frequency. If f is increasing then beat frequency has to decrease from 1st expression while it will increase in, accordance with 2nd expression. For given situation 1st equation is valid, so \[f=512-6=506Hz.\]You need to login to perform this action.
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