Question

Derive an expression for the stationary wave. Hence state the condition for nodes and anti nodes in a stationary wave.

Accepted Answer

Consider two transverse waves of the same wavelength, amplitude and frequency travelling along the +X-axis and -X-axis.

The wave travelling in +directions is given by,

$y' = a \sin (kx - ωt)$ ..................................................(1)

The wave travelling along the negative direction is.

$y'' = a \sin (kx + ωt)$ .............................................(2)

According to the principle of superposition, their combined wave is given by,

$Y = y' + y'' = a \sin (kx - ωt) + a \sin (kx + ωt)$

[As $sinA + sinB = 2 \sin (\frac{A + B}{2}) \cos (\frac{A - B}{2})$ }.

$Y = a [\sin (kx - ωt) + \sin (kx + ωt] = 2 a \sin (kx) \cos (- ωt)$ .

= $(2 a sinkx) \cos ωt$ . ...........(3) (As $\cos (- θ) = \cos θ$

The amplitude of the standing wave is $A = 2 \sin kx$ .................(4)

The amplitude will be maximum if $sinkx = 1 or kx = (n + \frac{1}{2}) π where n = 0.1.2.3 . . . . .$

Since $k = \frac{2 π}{λ} so k = (n + \frac{1}{2}) \frac{λ}{2}$ .., the maxima are at.

$x = \frac{λ}{4}, \frac{3 λ}{4}, \frac{5 λ}{4} . . .$ The position of maximum displacement is called antinodes.

For minimum amplitude, $\sin kx = 0 or kx = nπ where n = 0, 1, 2, 3, . .$

or $\frac{2 π}{λ} x = nπ or x = \frac{n}{2} λ$ . therefore, the minima or the nodes will be at $x = 0, \frac{λ}{2}, \frac{3 λ}{2} . . . . . .$ ,

It is clear from the above that the distance between successive nodes or antinodes is $\frac{λ}{2}$ .

Embibe Experts Solutions for Chapter: Solved Physics Paper-2, Exercise 1: Exercise - Choice

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