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Derive an expression for the equation of stationary wave on a stretched string. - Physics

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Derive an expression for the equation of stationary wave on a stretched string.

Explain the formulation of stationary waves by the analytical method.

Derivation
Explain

Solution

Consider two simple harmonic progressive waves of equal amplitudes (A) and wavelength (λ) propagating on a long uniform string in opposite directions (remember 2π/λ = k and 2πn = ω).

The equation of wave travelling along the x-axis in the positive direction is

`"y"_1 = a sin { 2π (nt - x/λ) }     ...(1)`

The equation of wave travelling along the x-axis in the negative direction is

`"y"_2 = a sin { 2π (nt + x/λ) }     ...(2)`

When these waves interfere, the resultant displacement of particles of string is given by the principle of superposition of waves as

y = y1 +y2

`y = a sin {2π (nt - x/λ)} + a sin {2π (nt + x/λ)}`

By using,

`sin "C"+sin"D"=2sin(("C"+"D")/2)cos(("C"-"D")/2)`, we get

 y = 2a sin (2πnt) cos `( 2πx)/λ`

y = 2a cos `(2πx)/λ` sin (2πnt) or,    ...(3)

Using 2a cos `(2πx)/λ` = A in equation 3, we get

y = A sin (2πnt)

As ω = 2πn, we get, y = A sin ωt.

This is the equation of a stationary wave, which gives resultant displacement due to two simple harmonic progressive waves. It may be noted that the terms in position x and time t appear separately and not as a combination 2π (nt ± x/λ).

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Stationary Waves
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Chapter 6: Superposition of Waves - Exercises [Page 156]

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Balbharati Physics [English] 12 Standard HSC Maharashtra State Board
Chapter 6 Superposition of Waves
Exercises | Q 5 | Page 156
SCERT Maharashtra Physics [English] 12 Standard HSC
Chapter 6 Superposition of Waves
Long Answer | Q 1. i.

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