Question

.A loop of a string of mass per unit length $\mu$ and radius R is rotated about an axis passing through the centre perpendicular to the plane with an angular velocity $\omega$ . A small disturbance is created in the loop having the same sense of the rotation. The linear speed of the disturbance for a stationary observer is

• A. $\omega R$
• B. $2\omega R$
• C. $3\omega R$
• D. $Zero$

Answer 1

Step 1: given data and diagram

For equilibrium of elementary mass

$dm.{\text{ω}}^{2}R=2Tsin\frac{d\theta }{2}$

Step 2: Calculation of The linear speed of the disturbance for a stationary observer is

Then $sind\theta \approx \frac{d\theta }{2}$

$=\mu Rd\theta {\omega }^{2}R=2T\frac{d\theta }{2}$

[∵ length of the segment$=Rd\theta$ and mass of segment$=\mu \times Rd\theta$ ]

$=\mu {\omega }^2{R}^2=T$

Hence, the velocity of wave w.r.t the string is

$V_w=\sqrt{\frac{T}{\mu }}=\sqrt{{\omega }^2{R}^2}=\omega R$

Also, the speed of the string is ωR

Therefore, the velocity of the disturbance w.r.t. ground $=\omega R+\omega R=2\omega R$

Hence the correct option is B