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A string is stretched between fixed points separated by 75 cm. It is observed to have resonant frequencies of 420 Hz and 315 Hz. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is

  • Option 1)

    10.5 Hz

  • Option 2)

    105 Hz

  • Option 3)

    1.05 Hz

  • Option 4)

    1050 Hz.


Answers (1)


As we learnt in

Vibration of composite string when joint is a node (N) -

\frac{n_{1}}{n_{2}}= \frac{l_{1}}{l_{2}}\sqrt{\frac{\mu _{1}}{\mu _{2}}}


- wherein

n_{1}\: and\: n_{2} are number of loops in two cases

l_{1}\: and\: l_{2} are length of two parts

\mu_{1}\: and\: \mu_{2} are linear density in two parts




Let the successive loops formed be P and (P+1) for frequencies 315 Hz and 420 Hz

\therefore \upsilon = \frac{p}{2l}\sqrt{\frac{T}{\mu }}= \frac{p\nu }{2l}

\therefore \frac{p\nu }{2l}= 315\: Hz\: and\: \frac{\left ( p+1 \right )\nu }{2l}=420Hz

or\: \: \: \frac{\left ( p+1 \right )\nu }{2l}-\frac{p\nu }{2l}=420- 315

\frac{\nu }{2l}=105\Rightarrow \frac{1\times \nu }{2l}= 105Hz

p = 1 for fundamental mode of vibration of string.

Lowest resonant frequency = 105 Hz.

Correct option is 2.


Option 1)

10.5 Hz

This is an incorrect option.

Option 2)

105 Hz

This is the correct option.

Option 3)

1.05 Hz

This is an incorrect option.

Option 4)

1050 Hz.

This is an incorrect option.

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