First slide

Travelling wave

Question

Two pulleys of negligible mass are connected to two blocks of mass m₁ and m₂ by two strings of the same material and the same cross-sectional area as shown in Fig. The speed of a transverse wave in string AB is v₁ and that in string CD is v₂. The ratio $\frac{v_{1}}{v_{2}}$ is

Moderate

A

1
B

2
C

$\sqrt{2}$
D

$\frac{1}{\sqrt{2}}$

Solution

Speed of transverse wave in a stretched string is $v = \sqrt{\frac{T}{μ}}$ where T is the tension in the string and $μ$ is the mass per unit length of the string. Since strings AB and CD have the same density and the same cross-sectional area, they have the same $μ$ . Hence $v \propto \sqrt{T}$
If the tension in CD is T, the tension in string AB is $\frac{T}{2} .$
$∴ \frac{v_{1}}{v_{2}} = \sqrt{\frac{T_{AB}}{T_{CD}}} = \sqrt{\frac{T / 2}{T}} = \frac{1}{\sqrt{2}}$
So the correct choice is (d).

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App