JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

# Wave Equation

An oscillation that travels through space accompanied by the transfer of energy is called a wave. The motion of the waves causes a transfer of energy without any displacement of the particles of the medium. The particles vibrate or oscillate at a fixed point. The mechanical waves travel through the medium, deforming the medium. The deformation of the medium will reverse itself due to the restoring force created during the deformation.

## Wave Equation

Consider a string kept at a constant tension, T, oscillating at an amplitude A. Let us consider a point P on the string. We can notice that the point P on the wave will oscillate with simple harmonic motion between y = A and y = -A with a time period T. The wave produced in SHM is sinusoidal. It means that the wave can be expressed in terms of sine function or cosine function with the appropriate value of amplitude, frequency and phase.

The general form of the sinusoidal wave is y (x,t) = A sin(kx – ωt + Φ)

Where,

A is the amplitude of the wave

k is the wavenumber, k = 2π/λ

ω is the angular frequency, ω = 2πf

Φ is the initial phase of the wave

(kx – ωt + Φ) is the total instantaneous phase

A phase change of 2nπ will give points oscillating in phase.

### Solved Examples

1. A transverse wave on a string is given by the equation y(x,t) = A sin (kx – ωt) = 0.2 m sin (6.28 m-1x – 1.57 s-1t). Find the amplitude, time period and speed of the wave.

Solution:

Amplitude, A = 0.2 m

Time period = 2π/ω

= (2 x 3.14)/1.57

= 4 seconds

Speed of the wave, v = ω/k

= 1.57/6.28

= 0.25 m/s

2. The equation of the wave given is y = 0.2 sin (0.04t + .02 x – π/6). The time is expressed in seconds and distance in cm. Find the minimum distance between two particles.

Solution:

Comparing the equation with y = A sin (ωt + kx – Φ)

k = 2π/λ = 0.02

λ = (2 x 3.14)/0.02 = 314 cm

3. The equation of the wave given by y = 0.002 sin 2π(5t – x/12), all the quantities are given in SI units. Calculate the displacement of the particle at a distance of 5 m from the origin after 0.2 s.

Solution

The equation given is y = 0.002 sin 2π(5t – x/12)

y = 0.002 sin 2π(5(0.2) – 5/12)

= 0.002 sin 2π(7/12)

= 0.002 sin 7π/6

= 0.002 sin (6π+π)/6

= 0.002 sin (π+π)/6

= – 0.002 sin (π/6)

y = – 0.002 x (½)

= – 0.001 m

Displacement of the particle = – 0.001 m

## Frequently Asked Questions on Wave Equation

Q1

### Define the crest of a wave.

The highest point in the region of elevation of a medium through which the wave travels is called the crest. The displacement of the particles of the medium at the crest is maximum and positive.

Q2

### Define the trough of a wave.

The lowest point in the region of depression of a medium is called the trough. The displacement of the particles of the medium at the trough is minimum and negative.

Q3

### Give two examples of transverse waves.

A wave in a stretched string.
Electromagnetic waves

Q4

### Define wavelength.

The distance between two successive crests or two successive troughs is called wavelength.