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QJEEAdv08Phf51f: JEE Advanced 2008 Physics - Class 11 JEEAdv08Phf51f Solution: A vibrating string of certain length 1 under a tension T…

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:

Multiple Choice Options

  • A 344
  • B 336
  • C 117.3
  • D 109.3

Step-by-Step Verified Solution

Let f be frequency of string

According to close open pipe, we have

⇒f=3V4l=3×3404×75×10−2

=340 Hz

Where, V=340 m/s

l=75 cm ×10−2 m

=75×10−2 m

And, n = frequency of tuning fork

=(340+4) or (340−4)

⇒n=344 or 336

As tension T increase, beat frequency decrease

But frequency of tuning fork increases

∴ n=344 Hz

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