JEE Advance - Physics (2008 - Paper 2 Offline - No. 8)
A vibrating string of certain length 1 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 per second when excited along 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 be 340 m/s, the frequency n of the tuning fork in Hz is:
344
336
117.3
109.3
Explanation
Let f be frequency of string
According to close open pipe, we have
$$ \Rightarrow f = {{3V} \over {4l}} = {{3 \times 340} \over {4 \times 75 \times {{10}^{ - 2}}}}$$
$$ = 340$$ Hz
Where, $$V = 340$$ m/s
$$l = 75$$ cm $$ \times {10^{ - 2}}$$ m
$$ = 75 \times {10^{ - 2}}$$ m
And, n = frequency of tuning fork
$$ = (340 + 4)$$ or $$(340 - 4)$$
$$ \Rightarrow n = 344$$ or 336
As tension T increase, beat frequency decrease
But frequency of tuning fork increases
$$\therefore$$ $$n = 344$$ Hz
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