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  • How to solve this problem- - Oscillations and Waves - JEE Main-2

A metal wire of linear mass density of  9.8 g/m is stretched with a tension of 10 kg -wt  between two rigid supports 1 metre apart. The wire passes at its middle point between the poles of a permanent magnet, and it vibrates in resonance when carrying an alternating current of frequency \upsilon  The frequency \upsilon of the alternating source is

  • Option 1)

    50 Hz

  • Option 2)

    100 Hz

  • Option 3)

    200 Hz

  • Option 4)

    25 Hz

 

Answers (1)

best_answer

As we learnt in

Standing wave in a string fixed at both ends -

 

- wherein

\nu _{n}= \frac{n}{2L}.\sqrt{\frac{T}{\mu }}

n= 1,2,3...........

\nu _{n} frequency of nth harmonic .

T= tension in string

\mu = mass / length

 

 

 

At resonance , frequency of vibration of wire become equal to frequency of  a.c.

For vibration of wire,\upsilon = \frac{1}{2l}\sqrt{\frac{T}{\mu }}

\therefore \upsilon = \frac{1}{2\times 1}\sqrt{\frac{10\times 9.8}{9.8\times 10^{-3}}}= \frac{100}{2}= 50Hz

Correct option is 1.


Option 1)

50 Hz

This is the correct option.

Option 2)

100 Hz

This is an incorrect option.

Option 3)

200 Hz

This is an incorrect option.

Option 4)

25 Hz

This is an incorrect option.

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