Tension of a stretched string is increased by 69% . In order to keep its frequency of vibration cons

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14.Waves and Sound

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The tension of a stretched string is increased by $69\%$. In order to keep its frequency of vibration constant, its length must be increased by ..... $\%$

A

$30$

B

$20$

C

$69$

D

$\sqrt {69 } $

Solution

$\mathrm{f}=\frac{\mathrm{p}}{2 l} \sqrt{\frac{\mathrm{T}}{\mathrm{m}}}$

$\frac{\mathrm{T}}{l^{2}}=\frac{4 \mathrm{f}^{2} \mathrm{m}}{\mathrm{p}^{2}}=$ constant

or $\ell^{2} \propto \mathrm{T} \quad$ or $\quad 1 \propto \mathrm{T}^{1 / 2}$

$\frac{l^{\prime}}{l}=\left(\frac{\mathrm{T}^{\prime}}{\mathrm{T}}\right)^{1 / 2}=\left(\frac{\mathrm{T}+0.69 \mathrm{T}}{\mathrm{T}}\right)^{1 / 2}$

$=(1.69)^{1 / 2}=1.3$

If the lenght is increased by $x \%$ then

${l^{\prime}=l+\frac{\mathrm{x} l}{100}}$

$\therefore$ ${\frac{l+\frac{\mathrm{x}}{100} l}{l}=1.3}$

$1+\frac{\mathrm{x}}{100} =1.3 $

$x/100 = 0.3$

or $\mathrm{x} =30 \%$

Standard 11

Physics

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