If the tension in a stretched string fixed at both ends is increased by $21$ %, the fundamental frequency is found to be changed by $15$ Hz. Then, the

original frequency is

150

Hz.

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velocity of propagation of the transverse wave along the string increases by

5

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velocity of propagation of the transverse wave along the string increases by

10

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fundamental wavelength on the string does not change

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Solution

The correct option is D fundamental wavelength on the string does not change
Since $v_{0} \propto \sqrt{T}$
$\frac{v_{0} + 15}{v_{0}} = \sqrt{\frac{1 + 0.21}{1}} = 1.1$
$\Rightarrow 15 = 0.1 v_{0} \Rightarrow v_{0} = 150 H z$
$\frac{v_{2}}{v_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{1.21}{1}} = 1.1$
$\Rightarrow v_{2} = 110$ % of $v_{1}$
$λ_{0} = 2 l$ , which does not change
So, (a), (c) and (d) are correct.

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If the tension in a stretched string fixed at both ends is increased by 21%, the fundamental frequency is found to be changed by 15 Hz. Then, the

The correct option is D fundamental wavelength on the string does not changeSince v0∝√T v0+15v0 = √1+0.211 = 1.1 ⇒ 15 = 0.1 v0 ⇒ v0 = 150 Hz v2v1 = √T2T1 = √1.211 = 1.1 ⇒ v2 = 110 % of v1 λ0 = 2l, which does not change So, (a), (c) and (d) are correct.

If the tension in a stretched string fixed at both ends is increased by $21$ %, the fundamental frequency is found to be changed by $15$ Hz. Then, the