Question

A rope of mass $m_0$ and length L is suspended vertically. If a mass m is suspended from the bottom of the rope, find the time for a transverse wave to travel the length of the rope.

• A. $2\sqrt{\frac{L}{m_{0}g}}(\sqrt{m_0+m}-m)$
• B. $\sqrt{\frac{L}{m_{0}g}}(\sqrt{m_0+m}-m)$
• C. $\sqrt{\frac{L}{2m_{0}g}}(\sqrt{m_0+m}-m)$
• D. $2\sqrt{\frac{L}{m_{0}g}}(\sqrt{m_0+m}+m)$

Answer 1

The correct option is A $2\sqrt{\frac{L}{m_{0}g}}(\sqrt{m_0+m}-m)$


At a distance x from the bottom tension is $(\frac{m_0}{L}\times g+mg)$
At this position speed of transverse wave $=\sqrt{\frac{(\frac{m_0}{L}x+m)g}{\left(\frac{m_0}{L}\right)}}$
$or\frac{dx}{dt}=\sqrt{\frac{L_g}{m_0}(\frac{m_{)}}{L}x+m)}$
$or{\int }_0^L{(\frac{m_{)}}{L}x+m)}^{-\frac{1}{2}}dx=\sqrt{\frac{L_g}{m_0}}{\int }_0^{t}dt$
$\frac{{\left[{(\frac{m_0}{L}x+m)}^{\frac{1}{2}}\right]}_0^L}{\frac{1}{2}.\frac{m_0}{L}}=\sqrt{\frac{L_g}{m_0}}t$
$\frac{2L}{m_0}[{(\frac{m_0}{L}.L+m)}^{\frac{1}{2}}-(0+m{)}^{\frac{1}{2}}]=\sqrt{\frac{L_g}{m_0}}t$
$\sqrt{m_0+m}-\sqrt{m}=\frac{1}{2}\sqrt{\frac{m_{0}g}{L}}t$
$t=2\sqrt{\frac{L}{m_{0}g}}(\sqrt{m_0+m}-\sqrt{m})$
Note: if m<< $m_0$, $t=2\sqrt{\frac{L}{g}}$