# Waves on a string

A string of mass per unit length $\mu$ is under tension $T$. The speed of a wave travelling on this string is given by \begin{align} v=\sqrt{\frac{T}{\mu}}. \end{align}

Two similar waves travelling in the opposite direction produces standing waves. The displacement of superposed wave is zero at the nodes and it is the maximum at the antinodes.

## Stationary waves on a string fixed at both ends

Consider a string of length $L$ fixed at the both ends. The wave displacement at the both ends is zero. This is given by boundary conditions: $y=0 \text{ at } x=0 \text{ and at } x=L$.

The frequencies allowed on this string are given by conditions \begin{align} L&=n\frac{\lambda}{2},\\ \nu&=\frac{n}{2L}\sqrt{\frac{T}{\mu}},\\ n&=1,2,3,\ldots. \end{align}

The fundamental or 1st harmonics is given by \begin{align} \nu_0=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \end{align}

1st overtone of 2nd harmonics is given by \begin{align} \nu_1=\frac{2}{2L}\sqrt{\frac{T}{\mu}} \end{align}

2nd overtone of 3rd harmonics is given by \begin{align} \nu_2=\frac{3}{2L}\sqrt{\frac{T}{\mu}} \end{align}

All harmonics are present in a string fixed at the both ends.

Sonometer is a string fixed at the both ends. Following are useful relations for a sonometer \begin{align} \nu & \propto \frac{1}{L} \\ \nu & \propto \sqrt{T} \\ \nu &\propto \frac{1}{\sqrt{\mu}} \\ \nu & =\frac{n}{2L}\sqrt{\frac{T}{\mu}} \end{align}

## Stationary waves on a string fixed at one end

Consider a string of length $L$ fixed at one end and free at the other end. The wave displacement at the fixed end is zero. This is given by boundary conditions: $y=0 \text{ at } x=0$.

The frequencies allowed on this string are given by conditions \begin{align} L & =(2n+1)\frac{\lambda}{4}\\ \nu &=\frac{2n+1}{4L}\sqrt{\frac{T}{\mu}} \\ n&=0,1,2,\ldots \end{align}

The fundamental or 1st harmonics is given by \begin{align} \nu_0=\frac{1}{4L}\sqrt{\frac{T}{\mu}} \end{align}

1st overtone of 3rd harmonics is given by \begin{align} \nu_1=\frac{3}{4L}\sqrt{\frac{T}{\mu}} \end{align}

2nd overtone of 5th harmonics is given by \begin{align} \nu_2=\frac{5}{4L}\sqrt{\frac{T}{\mu}} \end{align}

Only odd harmonics are present in a string fixed at one end.

## Questions

Question 1: A string is fixed at the two ends. The fundamental frequency is $\nu_0$. The string is plucked at the middle and then released. Mark the correct statement.

1. The string will vibrate in fundamental mode with a unique frequency $\nu_0$.
2. The vibration will be a superposition of modes with frequencies mainly $\nu_0, 2\nu_0, 3\nu_0$.
3. The vibration will be a superposition of modes with frequencies mainly $\nu_0, 3\nu_0, 5\nu_0$.

Question 2 The figure shown is an actual photograph of a string vibrating in its fundamental mode. Explain why the edges of the figure are sharper then the middle portion.

## Problems from IIT JEE

Problem (IIT JEE 2000): Two vibrating strings of the same material but of lengths $L$ and $2L$ have radii $2r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes, the one of length $L$ with frequency $\nu_1$ and the other with frequency $\nu_2$. The ratio $\nu_1/\nu_2$ is given by,

1. $2$
2. $4$
3. $8$
4. $1$

Solution: Let the lengths and radii of the two strings be $l_1=L$, $r_1=2r$, $l_2=2L$, and $r_2=r$, respectively. Let $\rho$ be the density of the material. The mass and mass per unit length of the first string are, \begin{align} m_1 &=\rho (\pi r_1^2 l_1) \\ &=4\pi \rho r^2 L,\\ \mu_1&=\frac{m_1}{l_1}=4\pi \rho r^2, \end{align} and that of the second string are, \begin{align} m_2&=\rho (\pi r_2^2 l_2) \\ &=2\pi \rho r^2 L, \\ \mu_2 & =\frac{m_2}{l_2}=\pi \rho r^2. \end{align} In fundamental mode, $l=\lambda/2=v/(2\nu)$ which gives, \begin{align} \nu=\frac{v}{2l}=\frac{1}{2l}\sqrt{\frac{T}{\mu}}. \end{align} Substitute $\mu$ from first and second equation to get, \begin{align} \nu_1&=\frac{1}{2l_1}\sqrt{\frac{T_1}{\mu_1}} \\ &=\frac{1}{2L}\sqrt{\frac{T}{4\pi\rho r^2}} \\ &=\frac{1}{4L}\sqrt{\frac{T}{\pi\rho r^2}},\nonumber\\ \nu_2 & =\frac{1}{2l_2}\sqrt{\frac{T_2}{\mu_2}} \\ &=\frac{1}{2(2L)}\sqrt{\frac{T}{\pi\rho r^2}} \\ &=\frac{1}{4L}\sqrt{\frac{T}{\pi\rho r^2}}.\nonumber \end{align}