question_answer

Two tuning forks when sounded together produced 4 beats/sec. The frequency of one fork is 256. The number of beats heard increases when the fork of frequency 256 is loaded with wax. The frequency of the other fork is [CPMT 1976; MP PMT 1993]

A)            504

B)            520

C)            260

D)            252

Correct Answer: C

Solution :

                   Suppose two tuning forks are named A and B with frequencies \[{{n}_{A}}=256\,Hz\] (known), nB = ? (unknown), and beat frequency x = 4 bps. Frequency of unknown tuning fork may be     \[{{n}_{B}}=256+4=260\,Hz\] or     \[=256-4=252\,Hz\] It is given that on sounding waxed fork A (fork of frequency 256 Hz) and fork B, number of beats (beat frequency) increases. It means that with decrease in frequency of A, the difference in new frequency of A and the frequency of B has increased. This is possible only when the frequency of A while decreasing is moving away from the frequency of B. This is possible only if nB = 260 Hz. Alternate method : It is given \[{{n}_{A}}=256\,Hz,\,{{n}_{B}}=?\] and x = 4 bps Also after loading A (i.e. nA ¯), beat frequency (i.e. x) increases (­). Apply these information?s in two possibilities to known the frequency of unknown tuning fork.           nA ¯ ? nB = x­        ... (i)           nB  ? nA ¯ = x­       ... (ii) It is obvious that equation (i) is wrong (ii) is correct so nB = nA + x = 256 + 4 = 260 Hz.