A string is stretched between fixed points separated by 75.0,, cm. It is observed to have reson

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14.Waves and Sound

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A string is stretched between fixed points separated by $75.0\,\, cm.$ It is observed to have resonant frequencies of $420\,\, Hz$ and $315\,\, Hz$. There are no other resonant frequencies between these two. The lowest resonant frequency for this string is .... $Hz$

A

$105$

B

$155$

C

$205$

D

$10.5$

(AIPMT-2015)

Solution

For a string fixed at both ends, the resonant frequencies are

$v_{n}=\frac{m v}{2 L}$ where $n=1,2,3, \ldots .$

The difference between two consecutive resonant frequencies is

$\Delta v_{n}=v_{n+1}-v_{n}=\frac{(n+1) v}{2 L}-\frac{n v}{2 L}=\frac{v}{2 L}$

which is also the lowest resonant frequency $(n=1)$

Thus the lowest resonant frequency for the given string

$=420 \mathrm{Hz}-315 \mathrm{Hz}=105 \mathrm{Hz}$

Standard 11

Physics

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