A uniform rope of length L and mass m_1 hangs | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Si

Vedclass · Accepted Answer

$\sqrt{\frac{m_1 + m_2}{m_2}}$

Vedclass · Answer

(A) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the linear mass density.Since the frequency $f$ of the pulse remains constant,the wavelength $\lambda = \frac{v}{f}$ is directly proportional to the velocity $v$. Therefore,$\lambda \propto \sqrt{T}$.At the lower end of the rope (where the block $m_2$ is attached),the tension $T_1$ is due to the weight of the block: $T_1 = m_2 g$.At the top of the rope,the tension $T_2$ is due to the weight of both the rope and the block: $T_2 = (m_1 + m_2) g$.Thus,the ratio of the wavelengths is:$\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{(m_1 + m_2) g}{m_2 g}} = \sqrt{\frac{m_1 + m_2}{m_2}}$.

Similar Questions

$A$ string of length $L$ is stretched by $\frac{L}{20}$ and the speed of transverse waves along it is $v$. The speed of the wave when it is stretched by $\frac{L}{10}$ will be (assume that Hooke's law is applicable).

The speed of a wave on a string is $150 \,ms^{-1}$ when the tension is $120 \,N$. The percentage increase in the tension in order to raise the wave speed by $20 \%$ is

$A$ string wave equation is given by $y=0.002 \sin (300 t-15 x)$ and the linear mass density is $\mu=0.1 \ kg/m$. Find the tension in the string (in $N$).

Obtain the equation of speed of transverse wave on a tensed (stretched) string.

The equation of a wave on a string of linear mass density $0.04 \ kg \ m^{-1}$ is given by $y = 0.02 \sin \left[ 2\pi \left( \frac{t}{0.04 \ s} - \frac{x}{0.50 \ m} \right) \right] \ m$. The tension in the string is .... $N$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module