IIT JEE Physics (1978 to 2018: 41 Years) Topic-wise Complete Solutions

# Longitudinal and Transverse Waves

## 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} \label{nab:eqn:1} 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} \label{nab:eqn:2} \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} \label{nab:eqn:3} \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 $y=a\sin(kx-\omega t)$ 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 $y=y_0 \sin\left(2\pi\left(ft-x/\lambda\right)\right)$. 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 $y=y_0 \sin\left(2\pi\left(ft-x/\lambda\right)\right)$ 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 $\lambda=\pi y_0/2$.