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 $\left[M L T^{-2}\right]$. 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 $d l$ and mass dm of string as shown in the below figure (a)

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

So For figure (b)

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

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

$$
\begin{aligned}
& \Rightarrow \frac{d m v^2}{R}=2 T \theta=T \frac{d l}{R} \\
& V^2=\frac{T}{d m / d l}
\end{aligned}
$$
Now using

$$
\mu=\frac{d m}{d l}
$$

we get

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