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Q.

A string of length 2L, obeying Hooke's law, is stretched so that its extension is L. The speed of the transverse wave travelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be

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a

$\frac{1}{2} υ$

b

$\frac{1}{\sqrt{2}} υ$

c

$\sqrt{2} υ$

d

$2 \sqrt{2} υ$

answer is D.

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Detailed Solution

$v = \sqrt{\frac{T}{μ}}$
1st Case $T = K Δ l = K l$
$ρ = \frac{m}{2 l + l} = \frac{m}{3 l}$
$v = \sqrt{\frac{3 K l^{2}}{m}}$
2nd Case $T = K Δ l = K (4 l)$
$ρ = \frac{m}{2 l + 4 l} = \frac{m}{6 l}$
$v^{1} = \sqrt{\frac{24 K l^{2}}{m}}$
$\frac{v^{1}}{v} = \sqrt{\frac{24}{3}} = \sqrt{8} = 2 \sqrt{2} v$
$T$ can be calculated by using Hooke’s law and on stretching $μ$ also changes.

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A string of length 2L, obeying Hooke's law, is stretched so that its extension is L. The speed of the transverse wave travelling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed of transverse wave travelling on the string will be