Question

A string of length 2L, obeying Hooke's Law, is stretched so that its extension is L. The speed of the transverse traveling on the string is v. If the string is further stretched so that the extension in the string becomes 4L. The speed transverse wave traveling on the string will be
A: $\sqrt{2}v$
B: $v$
C: $2\sqrt{2}v$
D: $2v$

Answer 1

Hint: Here, the string is extended twice to two different lengths. Hence, we can analyse the question in two different cases, that is when the elongation is up to L and when the elongation is up to 4L.

Complete step by step answer:
We know that, the speed of the transverse wave in given string $v=\sqrt{\dfrac{T}{\mu }}$
In case one,
$T=k\Delta L=kL$, since the string of 2L length extends to a length L
$\mu =\dfrac{m}{2L+L}=\dfrac{m}{3L}$
Hence, upon substitution, $v=\sqrt{\dfrac{3k{{L}^{2}}}{m}}$ (We are given that the value of the velocity is v)

In case two,
$T=k\Delta L=k(4L)$, since the string of 2L length extends to a length 4L
$\mu =\dfrac{m}{2L+4L}=\dfrac{m}{6L}$
Hence, upon substitution, ${{v}^{'}}=\sqrt{\dfrac{24k{{L}^{2}}}{m}}$
Comparing both velocities,
\[
 \dfrac{{{v}^{'}}}{v}=\sqrt{\dfrac{24}{3}}=\sqrt{8}=2\sqrt{2} \\
 \Rightarrow {{v}^{'}}=2\sqrt{2}v \\
\]
Hence, we can say that option C is the correct answer among the given options.

Note: Hooke’s law is used in most of the branches in science and it is the base of many disciplines including seismology, acoustics and molecular mechanics. It is also used for devices like a manometer, spring scale, and the balance wheel of the mechanical clock.