Question

A vibrating string of certain length `l` under a tension `T` resonates with a mode corresponding to the first overtone (third harmonic ) of an air column of length `75 cm` inside a tube closed at one end. The string also generates `4 beats//s` with a tuning fork of frequency `n` . Now when the tension of the string is slightly increased the number of beats reduces to `2` per second. Assuming the velocity of sound in air to `340 m//s` , the frequency `n` the tuning fork in `H_(Z)` is
(a) `344`
(b) `336`
(c ) `117.3`
(d) `109.3`
A. 344
B. 336
C. 117.3
D. 109.3

Answer 1

Correct Answer - A
With inc]rease in tension, frequency of vibrating string will increase. Since the number of beats are decreasing, therefore frequency of vibrating string or the third harmonic frequency of closed pipe should be less than the frequency of tuning fork by `4`.
Therefore, frequency of tuning fork = third harmonic frequency of closed pipe `+ 4`.
=`3((v)/(4l))+4=3((340)/(4 xx 0.75))+4 = 344 Hz`.