# Longitudinal and Transverse Waves

Mechanical waves are disturbances traveling in a medium (solid, liquid, or gas). The disturbance travels with a speed $v$ in a direction $\hat{k}$. The speed $v$ is called wave speed and the direction $\hat{k}$ is called the direction of wave propagation.

As the wave travels in a medium, the particles of the medium oscillate. If the direction of oscillation is along the direction of wave propagation then the wave is called longitudinal wave. If the direction of oscillation is perpendicular to the direction of wave propagation then the wave is called transverse wave.

Consider a long spring fixed at one end. If its other end is slightly pulled and released then the disturbance travels along its length. The particles of the spring oscillate along its length. It is a longitudinal wave.

If the spring is pulled aside and released then also disturbance travels along its length. In this case, particles of the spring oscillate perpendicular to its length. It is a transverse wave.

Consider the disturbace travelling on a string. The disturbance travels along the string but particles moves perpendicular to it. It is a transverse wave.

### Sound Waves

The disturbance for a sound wave travelling in air is pressure variation ($\Delta P$). The particles of the medium oscillates along the direction of propagation. It is a longitudinal wave.

Let us see particles of the medium at a particular instant. Particles are close to each other in certains regions (called compression) and far away from each other in other regions (called rarefaction). The pressure is high at compression and low at rarefaction.

## Problems from IIT JEE

Problem (IIT JEE 2005): A harmonically moving transverse wave on a string has a maximum particle velocity and acceleration of 3 m/s and ${90}\;\mathrm{m/s^2}$ respectively. Velocity of the wave is 20 m/s. Find the waveform.

Solution: The equation of a progressive wave travelling in the +$x$ direction is, \begin{align} y&=A\sin(kx-\omega t+\phi) \\ &=A\sin\left({2\pi x}/{\lambda}-2\pi\nu t+\phi\right). \end{align} Differentiate above equation w.r.t. time to get the particle velocity, \begin{align} \frac{\mathrm{d}y}{\mathrm{d}t}=-A\omega\cos(kx-\omega t+\phi), \end{align} which attains a maximum value of $\omega A$. Differentiate above equation w.r.t. time to get the particle acceleration, \begin{align} \frac{\mathrm{d}^2y}{\mathrm{d}t^2}=-A\omega^2\sin(kx-\omega t+\phi), \end{align} which attains a maximum value of $\omega^2 A$. Given, $\omega A={3}\;\mathrm{m/s}$ and $\omega^2 A={90}\;\mathrm{m/s^2}$. Solve to get $\omega={30}\;\mathrm{rad/s}$ and $A={0.1}\;\mathrm{m}$. The wave velocity $v=\nu\lambda={\omega}/{k}={20}\;\mathrm{m/s}$ gives $k=\omega/v={3/2}\;\mathrm{m^{-1}}$. The wave travelling towards the +$x$ direction is, \begin{align} y=0.1\sin\left({3x}/{2}-30t+\phi\right),\nonumber \end{align} and that travelling towards the $-x$ direction is, \begin{align} y=0.1\sin\left({3x}/{2}+30t+\phi\right).\nonumber \end{align}

Problem (IIT JEE 1999): In a wave motion $y=a\sin(kx-\omega t)$, $y$ can represent,

1. electric field
2. magnetic field
3. displacement
4. pressure

Solution: The equation $y=a\sin(kx-\omega t)$ represents a progressive wave traveling along the $x$ direction. The parameter $y$ is electric or/and magnetic field in electromagnetic waves, displacement of a particle in transverse waves on string, and pressure in the case of longitudinal sound waves.

Problem (IIT JEE 1984): A transverse wave is described by the equation \begin{align} y=y_0 \sin\left(2\pi\left(ft-x/\lambda\right)\right). \end{align} The maximum particle velocity is equal to four times the wave velocity if,

1. $\lambda=\pi y_0/4$
2. $\lambda=\pi y_0 /2$
3. $\lambda=\pi y_0$
4. $\lambda=2\pi y_0$

Solution: It can be seen that the wave \begin{align} y=y_0 \sin\left(2\pi\left(ft-x/\lambda\right)\right) \end{align} has the frequency $f$ and wavelength $\lambda$. Thus, the wave velocity is $v=f\lambda$.

The particle velocity for the given wave is, \begin{align} v_p&=\partial y/\partial t \\ &=y_0 (2\pi f) \cos\left(2\pi\left(ft-x/\lambda\right)\right).\nonumber \end{align} Thus, the maximum particle velocity is $v_{p,\text{max}}=2\pi f y_0$. The condition, $v_{p,\text{max}}=4v$, gives $\lambda=\pi y_0/2$.