Questions
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
detailed solution
Correct option is C
Suppose two tuning forks are named A and B with frequencies nA=256 Hz (known), nB = ? (unknown), and beat frequency x = 4 bps.Frequency of unknown tuning fork may benB=256+4=260 Hz or =256−4=252 HzIt 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 nA=256 Hz, nB=? and x = 4 bpsAlso after loading A (i.e. nA ↓), beat frequency (i.e. x) increases (↑).Apply these informations 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.Talk to our academic expert!
Similar Questions
A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance 50 cm the tension in the wire is 100 N the wire vibrating with its fundamental frequency and vibrating tuning fork together produces 5 beats per second. The tension in the wire is then reduced to 81 N, when the two are excited, beats are heard at the same rate. Calculate frequency of the fork.
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