If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

1 : 2

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\sqrt{2} : 1

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1 : 1

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1 : \sqrt{2}

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Solution

The correct option is D $1 : \sqrt{2}$
Velocity of transverse wave on a string

$v = \sqrt{\frac{T}{μ}}$

$T =$ Tension in the string

$μ =$ mass per unit length

$v_{1} = \sqrt{\frac{T_{1}}{μ}}$

$v_{2} = \sqrt{\frac{T_{2}}{μ}}$ ;

As per question $T_{2} = 2 T_{1}$

$\frac{v_{1}}{v_{2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

Hence, option $(A)$ is correct.

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If the initial tension on a stretched string is doubled, then the ratio of the initial and final speeds of a transverse wave along the string is:

The correct option is D 1:√2Velocity of transverse wave on a string v=√Tμ T= Tension in the string μ= mass per unit length v1=√T1μ v2=√T2μ; As per question T2=2T1 v1v2=√12=1√2 Hence, option (A) is correct.