(1) When two tuning forks are sounded together, 4 beats per second are produced. The frequency of one fork is 512Hz. When the other fork is loaded with a little wax, the beat frequency is 6Hz. The frequency of the other tuning fork is (in Hz)
(a) 516 (b) 514 (c) 512 (d) 508 (e) 502
This is a very simple question and you know that the beat frequency is the difference between the frequencies of the tuning forks. Since 4 beats are produced, the frequency of the other fork is either 516 or 508.
Loading of the other fork will reduce its frequency. So, if its actual frequncy is 516, the beat frequency is reduced. Since the beat frequency is increased, the actual frequency of the other fork must be 508 [option(d)].
Note that if the other fork is filed slightly, its frequency will be increased. If the question is modified this way and the beat frequency after filing is given as 6Hz itself, the answer will be option (a).
(2) The tension in a sonometer wire is increased by 96%. In order to keep the frequency of vibration of the wire unchanged, the length of the wire must be increased by
(a) 38% (b) 40% (c) 42% (d) 44% (e) 48%
The frequency of vibration of the wire is given by n = (1/2L)×√(T/m) where ‘L’ is the length, ‘T’ is the tension and and ‘m’ is the linear density of the wire. When T becomes 1.96T, the length L must become 1.4 L. The increment in L therefore is 40% [option (b)].
(a) 516 (b) 514 (c) 512 (d) 508 (e) 502
This is a very simple question and you know that the beat frequency is the difference between the frequencies of the tuning forks. Since 4 beats are produced, the frequency of the other fork is either 516 or 508.
Loading of the other fork will reduce its frequency. So, if its actual frequncy is 516, the beat frequency is reduced. Since the beat frequency is increased, the actual frequency of the other fork must be 508 [option(d)].
Note that if the other fork is filed slightly, its frequency will be increased. If the question is modified this way and the beat frequency after filing is given as 6Hz itself, the answer will be option (a).
(2) The tension in a sonometer wire is increased by 96%. In order to keep the frequency of vibration of the wire unchanged, the length of the wire must be increased by
(a) 38% (b) 40% (c) 42% (d) 44% (e) 48%
The frequency of vibration of the wire is given by n = (1/2L)×√(T/m) where ‘L’ is the length, ‘T’ is the tension and and ‘m’ is the linear density of the wire. When T becomes 1.96T, the length L must become 1.4 L. The increment in L therefore is 40% [option (b)].
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