Question

A flexible steel cable of total length \[\;L\] and mass per unit length $\mu $ hangs vertically from a support at one end. How long will it take for a wave to travel down the cable?
A. $t = 4\sqrt {\dfrac{L}{g}} $
B. $t = 4\sqrt {\dfrac{L}{{g - a}}} $
C. $t = 2\sqrt {\dfrac{L}{g}} $
D. $t = 9\sqrt {\dfrac{L}{{g + a}}} $

Answer 1

Hint: Use the relation between velocity and tension to find velocity. Then integrate it under the maximum limit of the length of the cable to find time taken by a wave to cover the entire length of the cable.

Formula Used: $T = \mu \left( {L - x} \right)g$
Where,
$T$ is tension in the rope
$\mu $ is mass per unit length
$L$ is the length of the cable
$x$ is a variable distance
$v( x ) = \sqrt {\dfrac{T}{\mu }}$
Where,
$v(x)$ is velocity at distance $x$

Complete step by step answer:Tension at a point at distance $x$ from the support is due to the weight of the cable below it.
Let $m$be the mass of the cable thus, $T = m\dfrac{{\left( {L - x} \right)}}{L}g$
Mass per unit length, $\mu = \dfrac{M}{L}$
Thus speed of the wave at given position $ = \sqrt { \dfrac{T}{\mu }} $
$ = \sqrt {\dfrac{{m\dfrac{{\left( {L - x} \right)}}{L}g}}{{\dfrac{m}{L}}}} $
$v = \sqrt {g\left( {L - x} \right)} $
Therefore, velocity of wave at distance $x$from support
$ = \sqrt {g\left( {L - x} \right)} $
$ \Rightarrow v = \dfrac{{dx}}{{dt}} = \sqrt {g\left( {L - x} \right)} $
By re-arranging the equation, we can write
$\dfrac{{dt}}{{dx}} = \dfrac{1}{{\sqrt {g\left( {L - x} \right)} }}$
Time taken to travers distance $dx$ is given by
$dt = \dfrac{{dx}}{{\sqrt {g\left( {L - x} \right)} }}$
To calculate the time taken to travel the entire length of the cable. We will integrate above equation from lower limit zero to upper limit $L$.
Thus, total time,
$t = \int\limits_0^L {\dfrac{{dx}}{{\sqrt {g\left( {L - x} \right)} }}} $
\[ = \dfrac{1}{{\sqrt g }}\int\limits_0^L {{{\left( {L - x} \right)}^{ - \dfrac{1}{2}}}dx} \]
We know that the integration of ${x^n}$ can be written as $\dfrac{{{x^{n + 1}}}}{{n + 1}}$
i.e. $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}} $
Therefore, the above equation can be written as
\[ = - \dfrac{1}{{\sqrt g }}\left[ {\dfrac{{{{\left( {L - x} \right)}^{ - \dfrac{1}{2} + 1}}}}{{ - \dfrac{1}{2} + 1}}} \right]_0^L\] . . . (1)
Simplifying it, we get
\[ = - \dfrac{1}{{\sqrt g }}\left[ {\dfrac{{{{\left( {L - x} \right)}^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}}} \right]_0^L\]
\[ = - \dfrac{2}{{\sqrt g }}\left[ {\sqrt {\left( {L - x} \right)} } \right]_0^L\]
Substituting the upper and lower limit of the length of the cable in the above equation, we get
$t = - \dfrac{2}{{\sqrt g }}\left[ {\sqrt {L - L} - \sqrt {L - 0} } \right]$
$ \Rightarrow t = 2\sqrt {\dfrac{L}{g}} $
Thus, total time taken for a wave to travel down the cable is $2\sqrt {\dfrac{L}{g}} $
Therefore, from the above explanation, the correct answer is, option (C) $t = 2\sqrt {\dfrac{L}{g}} $

Note:Information provided in this question is very less. We need to have a good understanding of different formulas to solve such questions. We needed a relation between distance and time, which is velocity. So, you can only solve this question, if you could relate tension to velocity. Knowing that formula is critical for this question.
If $\int {f(x)dx = F(x)} + C$ then $\int {f(g(x)) = \dfrac{{F(x)}}{{\dfrac{d}{{dx}}g(x)}}} + C$.
Therefore, the negative sign in equation (1) is because $\dfrac{d}{{dx}}(L - x) = - 1$