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← IIT-JEE 2008 · Paper 2
PhysicsSingle correct+3 / -1Question 29 of 66

IIT-JEE 2008 · Paper 2, Physics question 29

A vibrating string of certain length ℓ\ell 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 nn. 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 nn of the tuning fork in Hz is

Choose one option
Answer and worked solution

Official answer: A

Air column. Closed at one end, third harmonic: f=3v4L=3×3404×0.75=340f=\dfrac{3v}{4L}=\dfrac{3\times340}{4\times0.75}=340 Hz. The string vibrates at 340 Hz.

Fork. 4 beats give n=336n=336 or 344344 Hz. Increasing the tension raises the string's frequency; the beats drop to 2, so the string moved towards the fork: n=344n=344 Hz.

(A)