Question

Explain the formation of stationary waves.

Answer 1

Consider two harmonic progressive waves (formed by strings) that have the same amplitude and same velocity but move in opposite directions. Then the displacement of the first wave (incident wave) is

y₁ = A sin (kx – ωt) (waves move toward right) …(1)
and the displacement of the second wave (reflected wave) is

y₂ = A sin (kx + ωt) (waves move toward left) …(2)

both will interfere with each other by the principle of superposition, the net displacement is

y = y₁ + y₂ …… (3)

Substituting equation (1) and equation (2) in equation (3), we get

y = A sin (kx – ωt) + A sin (kx + ωt)   …(4)

Using trigonometric identity, we rewrite equation (4) as

y(x, t) = 2A cos (ωt) sin (kx) …(5)

This represents a stationary wave or standing wave, which means that this wave does not move either forward or backward, whereas progressive or travelling waves will move forward or backward. Further, the displacement of the particle in equation (5) can be written in more compact form,

y(x, t) = A’ cos (ωt)

where, A’ = 2A sin (foe), implying that the particular element of the string executes simple harmonic motion with amplitude equals to A’. The maximum of this amplitude occurs at positions for which

sin (kx) = 1 ${\displaystyle \Rightarrow \text{kx}=\frac{\pi }{2},\frac{3\pi }{2},\frac{5\pi }{2},...=\text{m}\pi }$

where m takes half-integer or half-integral values. The position of maximum amplitude is known as antinode. Expressing wave number in terms of wavelength, we can represent the anti-nodal positions as

${\displaystyle {\text{x}}_{\text{m}}=\left(\frac{2\text{m}+1}{2}\right)\frac{\lambda }{2}}$, where, m = 0, 1, 2...

For m = 0 we have maximum at ${\displaystyle x_0=\frac{\lambda }{2}}$

For m = 1 we have maximum at ${\displaystyle x_1=\frac{3\lambda }{4}}$

For m = 2 we have maximum at ${\displaystyle x_2=\frac{5\lambda }{4}}$ and so on.

The distance between two successive antinodes can be computed by

${\displaystyle {\text{x}}_{\text{m}}-{\text{x}}_{\text{m - 1}}=\left(\frac{2\text{m}+1}{2}\right)\frac{\lambda }{2}-\left(\frac{(2\text{m}+1)+1}{2}\right)\frac{\lambda }{2}=\frac{\lambda }{2}}$

Similarly, the minimum of the amplitude A’ also occurs at some points in the space, and these points can be determined by setting

sin (kx) = 0 ⇒ kx = 0, π, 2π, 3π, … = nπ

where n takes integer or integral values. Note that the elements at these points do not vibrate (not move), and the points are called nodes. The n^th nodal positions is given by,

${\displaystyle {\text{x}}_{\text{n}}=\text{n}\frac{\lambda }{2}}$ where, n = 0, 1, 2,...

For n = 0 we have minimum at ${\displaystyle x_0=0}$

For n = 1 we have minimum at ${\displaystyle {\text{x}}_1=\frac{\lambda }{2}}$

For n = 2 we have maximum at ${\displaystyle {\text{x}}_2=\lambda }$ and so on.

The distance between any two successive nodes can be calculated as

${\displaystyle {\text{x}}_{\text{n}}-{\text{x}}_{\text{n - 1}}=\text{n}\frac{\lambda }{2}-\left(\text{n - 1}\right)\frac{\lambda }{2}=\frac{\lambda }{2}}$