A string of length 1, m and mass 5, g is fixed at both ends. tension in string is 8.0, N . string is

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14.Waves and Sound

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A string of length $1\, m$ and mass $5\, g$ is fixed at both ends. The tension in the string is $8.0\, N$. The string is set into vibration using an external vibrator of frequency $100\, Hz$. The separation between successive nodes on the string is close to .... $cm$

A

$10$

B

$33.3$

C

$16.6$

D

$20.0$

(JEE MAIN-2019)

Solution

$\mathrm{f}=\frac{\mathrm{n}}{2 \ell} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

On solving, $n=5$

$5$ loops are formed in $1 \mathrm{m}$

Separation between successive nodes $=\frac{1}{5} \mathrm{m}=20 \mathrm{cm}$

Standard 11

Physics

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