Question

The separation between a node and an antinode in the stationary wave $y = 20\sin (2x)\cos 400\pi t$ is
(A) $\dfrac{\pi }{2}$
(B) $\dfrac{\pi }{4}$
(C) $\dfrac{\pi }{6}$
(D) $\dfrac{\pi }{8}$

Answer 1

Hint
Firstly, we will write down the formula for the superposition of two waves. We will compare the given equation with the known equation, $y = 2A\cos \dfrac{{2\pi x}}{\lambda }\sin 2\pi ft$ . Then we will write the value of frequency, resultant amplitude and the distance between node and antinode. That’s how we will get the solution.

Complete step by step answer
If two identical waves with equal amplitude, frequency, wavelength and velocity propagate through the same part of the medium simultaneously but in opposite directions the resultant wave is found to be confined in that part of the medium only. This resultant wave is called a stationary wave.
Let us consider two waves each having amplitude ‘A’ and velocity ‘v’.
So, the equation of the wave moving along positive x direction, ${y_1}$ [here $\lambda$ is the wavelength of the wave]
And the equation of the wave moving along negative x direction, ${y_2}$
From the principle of superposition, we will get,
After superposition of two waves we get, $y = 2A\sin \dfrac{{2\pi x}}{\lambda }\cos 2\pi ft.....(i)$
Frequency $(f) = \dfrac{v}{\lambda }$
According to the question, $y = 20\sin (2x)\cos 400\pi t...(ii)$ We know that the transverse wave consists of crests and troughs. We also know that distance between two successive crests or troughs is one wavelength $(\lambda )$ . The position where a crest and trough are formed and the particles vibrate with the amplitudes that are maximum are known as antinodes.
The distance between a crest and its successive trough is is half wavelength $\dfrac{\lambda }{2}$
So, the distance between two antinodes is $\dfrac{\lambda }{2}$ . The centre in between two adjacent antinodes there are particles that will be at rest and the positions of these particles is called nodes. They vibrate with minimum amplitude.
So, distance between nodes and antinodes will be half of the distance of half wavelength $= \dfrac{\lambda }{4}$
Comparing (i) and (ii) we get, $A = 10,\lambda = \pi ,f = 200Hz$
So, the separation between node and antinode will be
$\Rightarrow \dfrac{\lambda }{4} = \dfrac{\pi }{4}$
$\therefore \dfrac{\pi }{4}$ is the answer (option (B)).

Additional Information
All the particles in the same loop are situated in the same phase. The particles in consecutive loops are situated in opposite phases. Since the phase term is independent of space. Therefore, the phase is not propagating in space, that is the wave is called stationary.

Note
One might confuse that why the answer is $\dfrac{\pi }{4}$ not $\dfrac{\pi }{2}$ option (A). The length of the loop is $\dfrac{\lambda }{2}$ that is the distance between two consecutive nodes or antinodes will be $\dfrac{\lambda }{2}$ . So, the distance between the antinodes will be the half of the length of the loop. That’s why the separation is $\dfrac{\lambda }{2}$ .