Question

A wave pulse starts propagating in the x-direction along a non-uniform wire of length $20\text{m}$ with mass per unit length given by $\mu ={\mu }_0+ax$ and under a tension of $400\text{N}$. Find the time taken by the pulse to travel from the lighter end $(x=0)$ to the heavier end. ${\mu }_0={10}^{-2}\text{kg/m}$ and $a=16\times {10}^{-3}{\text{kg/m}}^2$

• A. 0.8 sec
• B. 0.4 sec
• C. 0.6 sec
• D. 1 sec

Answer 1

The correct option is B 0.4 sec

The speed of the wave pulse in a medium is given by
$V=\sqrt{\frac{T}{\mu }}=\sqrt{\frac{T}{{\mu }_0+ax}}=\frac{dx}{dt}$
To find the time taken for wave pulse to go from end to end, integrate above equation with limits of $x$ set from 0 to $L=20\text{m}$.
$\Rightarrow {\int }_0^t\sqrt{T}dt={\int }_0^L\sqrt{{\mu }_0+ax}dx$
$\Rightarrow \sqrt{T}[t{]}_0^t=\frac{2}{3a}\left[\right({\mu }_0+aL{)}^{\frac{3}{2}}-\left({\mu }_0{)}^{\frac{3}{2}}\right]$
$\Rightarrow t=\frac{2}{3\sqrt{T}a}\left[\right({\mu }_0+aL{)}^{\frac{3}{2}}-\left({\mu }_0{)}^{\frac{3}{2}}\right]$
$=\frac{2}{3\times \sqrt{400}\times 16\times {10}^{-3}}\left[\right({10}^{-2}+16\times {10}^{-3}\times 20{)}^{\frac{3}{2}}-\left({10}^{-2}{)}^{\frac{3}{2}}\right]$
$=2.08\times 0.188\simeq 0.4\text{s}$