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When two tuning forks  ( fork 1 and fork 2 ) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard. if the frequency of fork 1 is 200 Hz, then what was the original frequency of fork 2?

  • Option 1)

    196 Hz

  • Option 2)

    204 Hz

  • Option 3)

    200 Hz

  • Option 4)

    202 Hz


Answers (1)


As we learnt in

Property of stationary waves -

1)    Disturbance do not move in any direction.

2)    All particles except nodes perform S.H.M.

3)    During the formation of a stationary wave the medium is broken into equally spaced loops

4)    Amplitude of the particle is different at different point






Let the two frequencies\upsilon _{1} and \upsilon _{2}.\upsilon _{2} may be either 204 Hz or 196 Hz.

As mass of second fork increases, \upsilon _{2} decreases. If \upsilon _{2}= 204 Hz a decrease in \upsilon _{2} decreases beats/sec. But this is not given in question If \upsilon _{2}= 196 Hz a decrease in \upsilon _{2} increased beats/sec.

This is given in the question when beats increase to 6

Original frequency of second fork =196 Hz

Correct option is 1.

Option 1)

196 Hz

This is the correct option.

Option 2)

204 Hz

This is an incorrect option.

Option 3)

200 Hz

This is an incorrect option.

Option 4)

202 Hz

This is an incorrect option.

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