Question

Two transverse sine waves each of amplitude

4 mm

wavelength

2 m

and time period

1 s

and in phase at

x = 0, t = 0

are travelling along the

x

-axis in opposite direction. Obtain the equation of the resultant wave and comment on its nature calculate the maximum displacement at

x = 2.333 m .

Also locate the antinodes and nodes.

Accepted Answer

Given that two transverse waves are travelling along $x$ -axis in opposite direction and are in phase at $x = 0$ , $t = 0$ .

The amplitude of the wave $x = 0, t = 0$ , wavelength $λ = 2 m$ and time period $T = 1 s$ .

First wave equation is travelling along $+ x$ can be written as,

$y_{1} = A \sin (ω t - k x) y_{1} = A \sin (\frac{2 π}{T} t - \frac{2 π}{λ} x) We know k = \frac{2 π}{λ} and ω = \frac{2 π}{T}$

Putting the value of amplitude, wavelength and time period in the above equation. we get,

$y_{1} = 4 \times 10^{- 3} \sin (2 π t - \frac{2 π}{2} x) y_{1} = 4 \times 10^{- 3} \sin (2 π t - π x) \dots (1)$

Second wave equation is travelling along $- x$ can be written as,

$y_{2} = A \sin (ω t + k x) y_{2} = A \sin (\frac{2 π}{T} t + \frac{2 π}{λ} x)$

Putting the value of amplitude, wavelength and time period in the above equation. we get,

$y_{2} = 4 \times 10^{- 3} \sin (2 π t + \frac{2 π}{2} x) y_{2} = 4 \times 10^{- 3} \sin (2 π t + π x) \dots (2)$

The resultant wave equation wave can be written as.

$y_{R} = y_{1} + y_{2}$

$y_{R} = 4 \times 10^{- 3} [\sin (2 π t - π x) + \sin (2 π t + π x)]$

On simplifying. we get,

$y_{R} = 2 \times 4 \times 10^{- 3} \sin (\frac{2 π t - π x + 2 π t + π x}{2}) \cos (\frac{2 π t - π x - 2 π t - π x}{2}) y_{R} = 8 \times 10^{- 3} \cos (π x) \sin (2 π t)$

$y_{R} = 0$ at $x = 0, t = 0$

Amplitude of resultant wave is $8 \times 10^{- 3} \cos (π x)$

Maximum displacement at $x = 2.333$

$y_{R} = 8 \times 10^{- 3} \cos (2.333 π) y_{R} = 4 \times 10^{- 3} m$

For getting nodes. we can write,

$\cos (π x) = 0 \cos (π x) = \cos (2 n + 1) \frac{π}{2} π x = (2 n + 1) \frac{π}{2} Putting n = 0, 1, 2, 3 \dots we get x = \frac{1}{2}, \frac{3}{2} \dots$

For getting anti nodes. we can write,

$\cos (π x) = \pm 1 \cos (π x) = n π x = n Putting n = 1, 2, 3 we get anti nodes x = 1, 2, \dots$

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