Speed Of Transverse Wave On A String MCQ - Practice Questions with Answers

The distance between two successive crests is 1 wavelength, $\lambda$. Thus in one time period, the wave will travel 1 wavelength in distance. Thus the speed of the wave, $v$ is:

$$v=\frac{\lambda }{T}=\frac{\text{ Distance travelled }}{\text{ time taken }}$$

The speed of the traverse wave is determined by the restoring force set up in the medium when it is disturbed and the inertial properties ( mass density) of the medium. The inertial property will in this case be linear mass density $\mu$.
$\mu =\frac{m}{L}$ where m is the mass of the string and L is the length.
The dimension of ${\mu }_{\text{is }}\left[M{L}^{-1}\right]$ and T is like force whose dimension is $\left[ML{T}^{-2}\right]$. We need to combine these dimensions to get the dimension of speed v which is $\left[L{T}^{-1}\right]$.

Therefore speed of the wave in a string is given as:

$$v=\sqrt{\frac{T}{\mu }}$$

Now Let's understand its derivation.
Take a small element of length $dl$ and mass dm of string as shown in the below figure (a)

$$\text{ Here }dl=R(2\theta )$$

So For figure (b)

$$\frac{dm\times v^2}{R}=2T\sin \theta$$

For small $\theta$ we can use $\mathrm{Sin}\theta =\theta$

$$\begin{array}{rl}& \Rightarrow \frac{dm{v}^2}{R}=2T\theta =T\frac{dl}{R}\\ & V^2=\frac{T}{dm/dl}\end{array}$$
Now using

$$\mu =\frac{dm}{dl}$$

we get

$$\Rightarrow V=\sqrt{\frac{T}{\mu }}$$