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  4. A stretched string of length 1 m and mass 5 × 10-4 kg fixed at both ends is under tension of 20 N. it is plucked at a point situated at 25 cm from one end. The frequency of vibration of the string will be

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Q. A stretched string of length $1\, m$ and mass $5 \times 10^{-4} kg$ fixed at both ends is under tension of $20\, N$. it is plucked at a point situated at $25\, cm$ from one end. The frequency of vibration of the string will be

AMUAMU 2000

Solution:

The frequency $(n)$ of vibration of string is given by
$n=\frac{p}{2 l} \sqrt{\frac{T}{m}}$
where $T$ is tension, and $m$ the mass per unit length.
image
Given, $p=2, m=5 \times 10^{-4} kg$,
$T=20 \,N , l=1 \,m$
$\therefore n=\frac{2}{2 \times 1} \sqrt{\frac{20}{5 \times 10^{-4}}}$
$=200\, Hz$