# 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) Option 2) Option 3) Option 4)

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)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is the correct option.

Option 4)

This is an incorrect option.

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