Looking for classes? Ksquare Career Institute, Bengaluru →
A uniform rope of length L and mass $m_1$ hangs vertically from a rigid support. A block of mass $m_2$ is attached to the free end of the rope. A transverse pulse of wavelength $\lambda_1$ is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is $\lambda_2$. The ratio $\frac{\lambda_2}{\lambda_1}$ is
A
$\sqrt{\frac{m_1 + m_2}{m_2}}$
B
$\sqrt{\frac{m_2}{m_1}}$
C
$\sqrt{\frac{m_1 + m_2}{m_1}}$
D
$\sqrt{\frac{m_1}{m_2}}$
Explanation
Frequency is fixed and $\lambda \propto \sqrt{T}$.
Detailed Solution
Wave velocity $v = \sqrt{\frac{T}{\mu}}$ and $v = f\lambda$ with f constant, so $\lambda \propto \sqrt{T}$
Tension at the bottom $= m_2g$; tension at the top $= (m_1 + m_2)g$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2g}{(m_1 + m_2)g}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{m_1 + m_2}{m_2}}$
Tension at the bottom $= m_2g$; tension at the top $= (m_1 + m_2)g$
$\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{m_2g}{(m_1 + m_2)g}}$
$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{m_1 + m_2}{m_2}}$