Question

A wire of variable mass per unit length \[\mu = {\mu _0}x\] , is hanging from the ceiling as shown in figure. The length of wire is \[{l_0}\] . A small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach to the other end.

A.) \[\sqrt {\dfrac{{6{l_0}}}{g}} \]
B.) \[\sqrt {\dfrac{{2{l_0}}}{g}} \]
C.) \[\sqrt {\dfrac{{4{l_0}}}{g}} \]
D.) \[\sqrt {\dfrac{{8{l_0}}}{g}} \]

Answer 1

Hint: You can start by explaining what a transverse wave is. Then calculate the equation for the mass of wire $x$ distance from the lower end of wire by integrating the equation \[\mu = {\mu _0}x\] . Calculate the tension on the wire by using the equation $T = mg$ . Then calculate the speed of the transverse disturbance by using the equation Speed \[ = \sqrt {\dfrac{T}{m}} \] . Then calculate the time taken for the transverse disturbance to travel a small distance \[dx\] by using the equation Speed \[ = \dfrac{{Displacement}}{{Time}}\] and then integrate this equation to find the solution.

Complete step by step answer:

Transverse wave – Transverse wave or in this case a transverse disturbance is a wave in which the displacement of the particles of the medium is in a direction perpendicular to the direction of the propagation of the wave.

We are given that variable mass per unit length of the wire is given by the equation \[\mu = {\mu _0}x\] .So mass of the wire hanging at a distance \[x\] from the lower end of the wire (our reference point in this case) can be obtained by integrating the equation given above with respect to \[dx\]
\[\int\limits_0^x \mu dx = \int\limits_0^x {{\mu _0}x} dx\]

Mass of wire at a distance \[x\] is \[\int\limits_0^x \mu dx = {\mu _0}\dfrac{{{x^2}}}{2}\]
We know that there will be gravitational pull on the wire due to earth’s gravity and a tension will develop in the wire to counter this gravitational pull.
Gravitational pull on the wire \[ = mg = \] Tension on the wire \[ = {\mu _0}\dfrac{{{x^2}}}{2}g\]
We know that speed of a transverse is
Speed \[ = \sqrt {\dfrac{T}{m}} \]
Here, \[T = \] Tension and \[m = \] mass of the wire \[ = \mu \]
So Speed \[ = \sqrt {\dfrac{{{\mu _0}{x^2}g}}{{2\mu }}} \]
Speed \[ = \sqrt {\dfrac{{{\mu _0}{x^2}g}}{{2{\mu _0}x}}} \] ( \[\because \mu = {\mu _0}x\] )
Speed \[ = \sqrt {\dfrac{{gx}}{2}} \]
Let’s consider a small segment of the wire \[dx\] .
We know that Speed \[ = \dfrac{{Displacement}}{{Time}}\]
So the wire segment of length \[dx\] we get
\[\sqrt {\dfrac{{gx}}{2}} = \dfrac{{dx}}{T}\]
Here \[T = \] Time taken by the transverse disturbance to travel a distance \[dx\]
\[T = \dfrac{{dx}}{{\sqrt {\dfrac{{gx}}{2}} }}\]
So time taken to reach the other end of wire is
\[{T_0} = \int\limits_0^{{l_0}} {\dfrac{{dx}}{{\sqrt {\dfrac{{gx}}{2}} }}} = \sqrt {\dfrac{{8{l_0}}}{g}} \]
Hence, the correct answer is option D.

Note:
In the problem we are given a wire of variable mass per unit length, it does not mean that the wire is of non-uniform composition, it only means that the mass of wire which we consider will go on increasing as we move away from the point of reference, which in this case is the lower end of the wire.