Question

A string is hanging from a rigid support. A transverse pulse is excited at its free end. The speed at which the pulse travels a distance $x$ is proportional to​
A. $x$
B. $\dfrac{1}{x}$
C. $\dfrac{1}{{\sqrt x }}$
D. $\sqrt x $

Answer 1

Hint:To answer this question, one must know about transverse pulse, how to calculate its velocity. Then we can easily answer the question. Firstly, we will find out the velocity and then will put the value in the tension formula for the transverse pulse.

Formula used:
$V = \sqrt {\dfrac{T}{\mu }} $
Where, $V$ is the velocity, $T$ is the tension and $\mu $ is the linear mass density.

Complete step by step answer:
Let, the mass of the string is $m$, the length of the string be $L$, tension be $T$​ and $x$ be the distance travelled by the wave.
$V = \sqrt {\dfrac{T}{\mu }} \\
\Rightarrow T = {V^2}\mu \\ $
Also, we know that $V = \sqrt {xg} $ (the velocity $V$ of the pulse is related to the distance covered by it.)
Now putting the value of $V$ in equation $T = {V^2}\mu $ we get,
$T = xg\mu $
Again, putting the value of $T$ in $V = \sqrt {\dfrac{T}{\mu }} $ we get,
$V = \sqrt {\dfrac{{xg\mu }}{\mu }} $
From the above equation we can say that the speed at which the pulse travels a distance $x$ is proportional to​ $\sqrt x $.

Hence, the correct option is D.

Note:Don’t get confused in $V = \sqrt {xg} $ and $V = \sqrt {\dfrac{T}{\mu }} $ . The speed of a wave on this kind of string is proportional to the square root of the tension in the string and inversely proportional to the square root of the linear density of the string