Question

Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed `v_A` and on B with speed `v_B`. The ratio `v_A/v_B` is ______.

Options

• `1/2`

• 2

• `1/4`

• 4

Answer 1

Two strings A and B, made of same material, are stretched by same tension. The radius of string A is double of the radius of B. A transverse wave travels on A with speed `v_A` and on B with speed `v_B`. The ratio `v_A/v_B` is `underlinebb(1/2)`.

Explanation:

Wave speed is given by
\[\nu = \sqrt{\frac{T}{\mathrm{\mu}}}\]
where
T is the tension in the string
v is the speed of the wave
μ is the mass per unit length of the string

\[\mathrm{\mu}  = \frac{M}{L} = \rho\frac{V}{L} = \rho\frac{\left( AL \right)}{L}\]
where
M is the mass of the string, which can be written as ρV.

L is the length of the string.

\[= \rho\left( \pi r^2 \right) = \rho\left( \pi\frac{D^2}{4} \right)\] 

\[ \therefore \nu = \sqrt{\frac{T}{\rho\pi\frac{D^2}{4}}} = \frac{2}{D}\sqrt{\frac{T}{\rho\pi}}\]

where D is the diameter of the string.

Thus, v ∝
\[\frac{1}{D}\] Since, rA = 2rB 

\[v_A  \propto \frac{1}{2 r_A} \propto \frac{1}{2 \times 2 r_B}                                            (1)\] 

\[ v_{{}_B}  \propto \frac{1}{2 r_{{}_B}}                                                                            (2)\]

From Equations (1) and (2) we get  \[\frac{v_A}{v_B} = \frac{1}{2}\].