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IIT JEE Advanced 2008 Paper 2

Section: Physics

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Question : 2 of 66

Marks: +1, -0

A vibrating string of certain length l under 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

[JEE Adv 2008 P2]

344
336
117.3
109.3

Solution:

The third harmonic of the closed pipe is

f =

\frac{3V}{4l}

=

\frac{3 \times 340}{4 \times 0.75}

= 340 Hz

This is also the frequency of the string since γ of the string is proportional to

\sqrt{T}

. On increasing the tension, γ increases, and the number of beats decreases, given that n – 340 = 4. Therefore,

n > 340 Hz ⇒ n = 344 Hz

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