The tension of a stretched string is increase | vedclass

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(A) The fundamental frequency of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$l$ is the length,and $

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$30$

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(A) The fundamental frequency of a stretched string is given by $f = \frac{1}{2l} \sqrt{\frac{T}{m}}$,where $T$ is the tension,$l$ is the length,and $m$ is the mass per unit length.For a constant frequency $f$ and constant mass per unit length $m$,we have $\frac{1}{l} \sqrt{T} = 	ext{constant}$,which implies $l \propto \sqrt{T}$.Given that the tension is increased by $69\%$,the new tension $T' = T + 0.69T = 1.69T$.Let the new length be $l'$. Then $\frac{l'}{l} = \sqrt{\frac{T'}{T}} = \sqrt{1.69} = 1.3$.This means $l' = 1.3l$.The percentage increase in length is $\frac{l' - l}{l} 	imes 100 = \frac{1.3l - l}{l} 	imes 100 = 0.3 	imes 100 = 30\%$.

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 ..... $\%$

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