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NEET PHYSICSEasy

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:

A

1:21:2

B

1:11:1

C

2:1\sqrt{2}:1

D

1:21:\sqrt{2}

Step-by-Step Solution

The speed of a transverse wave on a stretched string is given by the formula v=Tμv = \sqrt{\frac{T}{\mu}}, where TT is the tension in the string and μ\mu is its linear mass density. Let the initial tension be T1=TT_1 = T, then the initial speed is v1=Tμv_1 = \sqrt{\frac{T}{\mu}}. If the tension is doubled, the new tension becomes T2=2TT_2 = 2T. The final speed is then v2=2Tμ=2Tμ=2v1v_2 = \sqrt{\frac{2T}{\mu}} = \sqrt{2} \sqrt{\frac{T}{\mu}} = \sqrt{2} v_1. The ratio of the initial speed to the final speed is v1v2=12\frac{v_1}{v_2} = \frac{1}{\sqrt{2}}, which can be written as 1:21:\sqrt{2}.

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