Wave Mechanics Question 110
Question: 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]
Options:
A) 504
B) 520
C) 260
D) 252
Show Answer
Answer:
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.
BETA
