A tuning fork of known frequency 256 Hz makes 5 beats per second with the vibrating string of a piano. The beat frequency decreases to 2 beats per second when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

  • Option 1)

    (256+2)Hz

  • Option 2)

    (256-2)Hz

  • Option 3)

    (256-5)Hz

  • Option 4)

    (256+5)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

 

-

 

 

 

The possible frequencies of piano are

\left ( 256+5 \right )Hz \: and \: \left ( 256-5 \right )Hz

For piano string, \upsilon = \frac{1}{2l}\sqrt{\frac{T}{\mu }}

When tension  T increases, \upsilon increases

(i) If  261 Hz increases, beats/ sec increase. This is not given in the question.

(ii)If 251 Hz increases due to tension, beats per second decrease. This is given in the question.

Hence frequency of piano= \left ( 256-5 \right )Hz

Correct option is 3.


Option 1)

(256+2)Hz

This is an incorrect option.

Option 2)

(256-2)Hz

This is an incorrect option.

Option 3)

(256-5)Hz

This is the correct option.

Option 4)

(256+5)Hz

This is an incorrect option.

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