Question

A standing wave set up in a medium, the equation of wave is given by

y = 4 \cos \frac{π x}{3} \sin 40 π t

where

x

and

y

are in

cm

and

t

in

\sec .

Find the amplitude and the velocity of the two component waves.

Accepted Answer

The given equation is: $y = 4 \cos \frac{π x}{3} \sin 40 π t$

or $y = 2 \times 2 \cos \frac{2 π x}{6} \sin \frac{2 π (120) t}{6} . . . (i)$

When two travelling waves $y_{1} = A \sin (ω t - k x) and y_{2} = A \sin (ω t + k x)$ interfere, the standard equation of a stationary wave is given by,

$y = 2 A \cos k x \sin ω t$

$\Rightarrow y = 2 Acos \frac{2 πx}{λ} \sin \frac{2 πvt}{λ} . . . (ii)$

where $k = \frac{2 π}{λ}$ is the propagation constant, $ω = k v = \frac{2 π}{λ} v$ is the frequency of the wave, $v$ is the velocity of the wave.

Comparing the Eq. $(i)$ and $(ii),$ we get

Amplitude, $A = 2 cm,$ Velocity, $v = 120 cm / s$ , and $λ = 6 cm$

The component waves are,

$y_{1} = A \sin \frac{2 π}{λ} (v t - x) = 2 \sin \frac{π}{3} (120 t - x)$

and $y_{2} = A \sin \frac{2 π}{λ} (v t + x) = 2 \sin \frac{π}{3} (120 t + x)$

B M Sharma Solutions for Chapter: Superposition and Standing Waves, Exercise 1: ILLUSTRATIONS

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Superposition and Standing Waves, Exercise 1: ILLUSTRATIONS

Questions from B M Sharma Solutions for Chapter: Superposition and Standing Waves, Exercise 1: ILLUSTRATIONS with Hints & Solutions

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