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 ${\nu }_A$ and on B with speed ${\nu }_B$. The ratio ${\nu }_A/{\nu }_B$ is
(a) 1/2
(b) 2
(c) 1/4
(d) 4.

Answer 1

(a) 1/2

Wave speed is given by
$\nu =\sqrt{\frac{T}{\mathit{\mu }}}$
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
$\mu =\frac{M}{L}=\rho \frac{V}{L}=\rho \frac{\left(AL\right)}{L}$
where
M is the mass of the string, which can be written as ρV
L is the length of the string
$$\begin{gathered} =\rho \left(\pi r^2\right)=\rho \left(\pi \frac{D^2}{4}\right) \\ \therefore \nu =\sqrt{\frac{T}{\rho \pi {\displaystyle \frac{D^2}{4}}}}=\frac{2}{D}\sqrt{\frac{T}{\rho \pi }} \end{gathered}$$
where D is the diameter of the string.
Thus, v ∝ $\frac{1}{D}$
Since, r_A = 2r_B
$$\begin{gathered} v_{\mathrm{A}}\propto \frac{1}{2r_{\mathrm{A}}}\propto \frac{1}{2\times 2r_{\mathrm{B}}}\left(1\right) \\ {v}_{{}_{\mathrm{B}}}\propto \frac{1}{2r_{{}_{\mathrm{B}}}}\left(2\right) \end{gathered}$$
From Equations (1) and (2) we get
$\frac{v_{\mathrm{A}}}{v_{\mathrm{B}}}=\frac{1}{2}$