A tuning fork vibrating with a sonometer havi | vedclass

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Question

The frequency of vibrations of a sonometer wire is given by :

$\mathrm{n}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

The tunin

Vedclass · Accepted Answer

$205$

Vedclass · Answer

The frequency of vibrations of a sonometer wire is given by :

$\mathrm{n}=\frac{1}{2 \mathrm{L}} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

The tuning fork produces $5$ beats per second with lengths $20 \mathrm{\,cm}$ and $21 \mathrm{\,cm}$. If $\mathrm{n}$ be frequency of fork, then

$\mathrm{n}+5=\frac{1}{2 	imes 0.20} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$          .........$(i)$

$\mathrm{n}-5=\frac{1}{2 	imes 0.21} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$           .........$(ii)$

Dividing eqn. $(i)$ by $(ii)$ we get:

hence, $\quad \frac{n+5}{n-5}=\frac{0.21}{0.20}$

or  $0.2 n+1=0.21 n-1.05$

or  $1+1.05=0.21 n-0.2 n^{\prime}$

or  $0.05=0.01\, n$

or  $\mathrm{n}=205 \mathrm{\,Hz}$

A tuning fork vibrating with a sonometer having $20\,cm$ wire produces $5$ beats per sec. The beat frequency does not change if the length of the wire is changed to $21\,cm$. The frequency of the tuning fork must be ..... $Hz$

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