A stretched string of $1m$ length and mass $5 \times {10^{ - 4}}kg$ is having tension of $20N.$ If it is plucked at $25cm$ from one end then it will vibrate with frequency ... $Hz$
Medium
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(b) As we know plucking distance from one end $ = \frac{l}{{2p}}$
==> $25 = \frac{{100}}{{2p}}$ ==> $p = 2$. Hence frequency of vibration
$n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} = \frac{2}{{2 \times 1}}\sqrt {\frac{{20}}{{5 \times {{10}^{ - 4}}}}} $ = $200Hz.$
==> $25 = \frac{{100}}{{2p}}$ ==> $p = 2$. Hence frequency of vibration
$n = \frac{p}{{2l}}\sqrt {\frac{T}{m}} = \frac{2}{{2 \times 1}}\sqrt {\frac{{20}}{{5 \times {{10}^{ - 4}}}}} $ = $200Hz.$
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