Question

A progressive wave is represented by $y = 12\sin \left( {5t - 4x} \right){\text{cm}}$. Find the distance between two points on this wave with a phase difference of $90^\circ$.
A) $\dfrac{\pi }{2}{\text{cm}}$
B) $\dfrac{\pi }{4}{\text{cm}}$
C) $\dfrac{\pi }{8}{\text{cm}}$
D) $\dfrac{\pi }{{16}}{\text{cm}}$

Answer 1

The argument of the sine function $\left( {5t - 4x} \right)$ denotes the phase of the wave and it is given as $90^\circ$ for two points at positions ${x_1}$ and ${x_2}$ at an instant $t$.

Formulae used:
The equation of a progressive wave is given by, $y\left( {x,t} \right) = a\sin \left( {kx - \omega t + \phi } \right)$
where $y\left( {x,t} \right)$ denotes the displacement from the equilibrium position along the y-axis, $x$ denotes the position of the propagating particles at a time $t$ , $a$represents the amplitude of the wave, $k$ is the wavenumber, $\omega$ is the angular frequency and $\phi$ is the initial phase of the wave.

Complete step by step answer:
Sketch the wave equation $y = 12\sin \left( {5t - 4x} \right)$ roughly.
The below figure roughly represents the displacement $y$ of the wave for various positions $x$ .
Define a progressive wave
A wave that continuously travels in the same direction without a change in its amplitude is called a progressive wave or travelling wave and its general form is given as, $y\left( {x,t} \right) = a\sin \left( {kx - \omega t + \phi } \right)$
Here, $y\left( {x,t} \right)$ denotes the displacement from the equilibrium position along the y-axis
The position of the propagating particles at a time $t$ is $x$.
The amplitude of the wave is represented by $a$ . It is the magnitude of maximum displacement of a particle from the equilibrium position.
The argument $\left( {kx - \omega t + \phi } \right)$ of the sine term is the phase of the function. It describes how the points on the wave rise and fall.
The angular frequency $\omega$ refers to the angular displacement of any particle of the wave per unit time.
The wavenumber $k$ denotes the number of waves that exist in a specified distance. A wavelength $\lambda$ corresponds to the distance between a rise or crest and a fall or trough.
and $\phi$ is the initial phase of the wave which suggests where the wave must start.
List the information provided by comparing the equation of the wave from the question $y = 12\sin \left( {5t - 4x} \right)$ with the general form $y\left( {x,t} \right) = a\sin \left( {kx - \omega t + \phi } \right)$.
The equation of the progressive wave is given as $y = 12\sin \left( {5t - 4x} \right){\text{cm}}$
Here, the amplitude of the wave is $a = 12{\text{cm}}$
On comparing, $k = 4$ is the wavenumber
Also, angular frequency $\omega = 5{\text{rad/s}}$
and the initial phase of the wave, $\phi = 0$
Consider two points on the wave to find the required distance between them
At an instant $t$ , let ${x_1}$ and ${x_2}$ be the positions of two points having a phase difference of $90^\circ = \dfrac{\pi }{2}$ on the wave.
Since the argument of the sine function, $\left( {5t - 4x} \right)$ denotes the phase of the wave, we can write, $\left( {5t - 4{x_1}} \right) - \left( {5t - 4{x_2}} \right) = \dfrac{\pi }{2}$
Simplifying, $5t - 4{x_1} - 5t + 4{x_2} - = \dfrac{\pi }{2}$
i.e., $4\left( {{x_2} - {x_1}} \right) = \dfrac{\pi }{2}$
Now, the distance between the two points is $\left( {{x_2} - {x_1}} \right) = \dfrac{\pi }{8}{\text{cm}}$

Therefore, the correct option is (C), $\dfrac{\pi }{8}{\text{cm}}$.

Note: Alternate method
The distance between the two points can be obtained using a relation given by,
$\phi = \dfrac{{2\pi }}{\lambda }d$ ---------- (A) where $\phi$ is the phase difference between the two points, $\lambda = \dfrac{{2\pi }}{k}$ is the wavelength of the wave and $d = {x_2} - {x_1}$ is the distance between the two points.
Since the wavenumber $k = 4$ , the wavelength will be $\lambda = \dfrac{{2\pi }}{4} = \dfrac{\pi }{2}{\text{cm}}$ . Given, phase difference is $\phi = \dfrac{\pi }{2}$ .
Then substituting for $\lambda = \dfrac{\pi }{2}{\text{cm}}$ and $\phi = \dfrac{\pi }{2}$ in equation (A) we get, $\dfrac{\pi }{2} = \dfrac{{4\pi }}{\pi }d$
Cancelling the similar terms and rearranging the above expression we get $d = \dfrac{\pi }{{2 \times 4}} = \dfrac{\pi }{8}{\text{cm}}$
Thus the distance between the two points is $d = \dfrac{\pi }{8}{\text{cm}}$.