What is a Wavelength? Definition and Solved Examples

Physics Topics can help us understand the behavior of the natural world around us.

What is the Relationship Between Wave and Wavelength?

Other Classifications of Waves

Considering the nature of vibration, we classify waves as mechanical waves, electromagnetic waves, etc. Again, considering the direction of motion of the vibrating particles of the medium and that of the wave, we classify them as transverse waves and longitudinal waves. In addition, waves can also be categorised on the basis of their different properties. A few of them are discussed below.

On the basis of direction of energy transmission: The wave which transmits energy in a single direction is called a one-dimensional wave e.g., transverse wave formed in a stretched string and longitudinal wave formed in an elastic spring are one-dimensional waves.

The wave which transmits energy along a plane is called a two-dimensional wave. The wave formed on the surface of water, when a stone is thrown on it, its two-dimensional.

The wave which transmits energy in all directions is called a three-dimensional wave. Sound waves, light waves, radio waves, etc., are three-dimensional waves.

On the basis of characteristics of particle vibration: If the particles of a medium vibrate simple harmonically, the corresponding wave is called simple harmonic wave. Practically most waves are produced by complex vibrations. But any complex wave may be described as superposition of a number of simple harmonic waves.

On the basis of limit of wave motion: If any wave advances through a medium continuously with a definite velocity, the wave is called a travelling or progressive wave. If the wave is not damped, it can propagate up to infinity. On the other hand, if the wave does not advance, but remains confined in a region, it is called a standing or stationary wave.

Some Physical Terms Related To Waves

Phase: The quantity from which the motion of a wave can be known completely is called the phase of the wave. Displacement, velocity, acceleration, etc., of a vibrating particle can be obtained from it. In Fig., the particles at O and D are in the same phase. The particles at A and E in Fig., or the particles at C and E in Fig., are also in the same phase. In Fig. particles at A and C and in Fig. particles at C and D are in opposite phases.

Complete wave: The wave in between two consecutive particles, having the same phase at an Instant, is known as a complete wave. Fig. and Fig. show respectively how a complete transverse wave is formed in between two consecutive crests, and a complete longitudinal wave is formed by the combination of a compression and a rarefaction.

Wavelength: The length of a complete wave, i.e., the distance between two consecutive particles having the same phase at an instant, is called the wavelength (λ). In Fig., OD and AE is the wavelength of the transverse wave. Again in Fig., CE and DF is the wavelength of the longitudinal wave. Generally it can be said that:

wavelength of a transverse wave = distance between any two consecutive crests or consecutive troughs.

wavelength of a longitudinal wave = total length of a pair of successive compression and rarefaction.

In general wavelength of a wave is the distance between two consecutive points in the same phase of motion at the same instance of time.

Time period: Time required to form a complete wave, i.e., time taken by a wave to cover a distance between two consecutive particles having the same phase of vibration, is called the time period (T) of the wave.

Frequency: Frequency (n) of a wave is the number of complete waves formed In unit time.

Amplitude: The amplitude of a wave is the maximum displacement of any particle producing the wave, from its mean position. In Fig., the distance of the points A, C or E from the straight line OBDF is the amplitude of the wave.

Wave velocity: Distance travelled by a wave in unit time is called its wave velocity (V). Energy is transmitted through the medium by the wave with this velocity. It may be noted that wave velocity is different from particle velocity. Particle velocity is the velocity of the particles of the medium which execute simple harmonic motions about their mean positions.

Wavefront: All the particles on a surface normal to the direction of propagation of a wave have the same phase. A surface of this type is called a wavefront. In other words wavefront is the locus of all points having the same phase of motion at the same instance of time. This surface is perpendicular to the direction of propagation of wave at any point. In Fig., the plane A is a wavefront, because the phase of all the particles lying on plane A is the same.

Similarly, the plane B is another wavefront. Clearly, planes A and B are parallel.

Definition: Any surface, which is normal to the direction of propagation of a wave, is known as a wavefront. The particles lying on a wavefront have the same phase.

Ray: A normal drawn on a wavefront is called a ray. The energy of a wave is transferred from one part of the medium to another along the ray.

Relations among the Physic Quantities

Relation between time period and frequency: If T is the time period of a wave, one complete wave is formed in time T. Therefore, the number of complete waves formed in unit time is \(\frac{1}{T}\).
So, according to the definition of frequency,
n = \(\frac{1}{T}\) or, T = \(\frac{1}{n}\) …… (1)

Relation between wavelength and wave number:

The number of complete waves in a length λ is 1.

So, the number of complete waves in unit length is \(\frac{1}{\lambda}\). This quantity multiplied by 2π is known as the wave number k,
k = 2π ᐧ \(\frac{1}{\lambda}\) = \(\frac{2 \pi}{\lambda}\) or, λ = \(\frac{2 \pi}{k}\) …… (2)

Relation among wave velocity, frequency and wavelength: Let the frequency of a wave be n, wavelength λ and wave velocity V. According to the definition of frequency, the number of complete waves formed in unit time is n. Again, the wave covers a distance V in unit time. So, the length of n number of complete waves is V.
So, the length of one complete wave = \(\frac{V}{n}\);
then, according to the definition of wavelength,
λ = \(\frac{V}{n}\) or, V = nλ …… (3)
i.e., wave velocity = frequency × wavelength
Again if T is the time period of a wave, n = \(\frac{1}{T}\)
∴ V = nλ = \(\frac{\lambda}{T}\) or, λ = VT ………. (4)
From equation (3), n = \(\frac{V}{\lambda}\). If more than one waves move through a medium with the same velocity, V will be a constant. In that case, n ∝ \(\frac{1}{\lambda}\), i.e., frequency and wavelength will be inversely proportional to each other. So, greater the wavelength of a wave, smaller its frequency and vice versa.

Numerical Examples

Example 1.
The wave generated on a water surface advances 1 m in 1 s. If the wavelength is 20 cm, how many waves are produced per second?
Solution:
Wavelength, λ = 20 cm;
wave velocity, V = 1 m ᐧ s^-1 = 1oo cm ᐧ s^-1
If n is the frequency, V = nλ
or, n = \(\frac{V}{\lambda}\) = \(\frac{100}{20}\) = 5 s^-1
So, 5 waves are produced per second.

Example 2.
A radio centre broadcasts radio waves of length 300 m. What is the frequency of this wave? Given, velocity of light = 3 × 10⁸ km ᐧ s^-1.
Solution:
Wavelength of radio wave, λ = 300 m.
Radio waves are electromagnetic waves like light and both of them have the same velocity.
So, the velocity of radio wave,
V = 3 × 10⁵ km ᐧ s^-1 = 3 × 10⁸ m ᐧ s^-1
∴ Frequency, n = \(\frac{V}{\lambda}\) = \(\frac{V}{\lambda}\) = 10⁶ Hz = 1 MHz.

Example 3.
The frequency of a tuning fork is 400 Hz and the velocity of sound in air is 320 m ᐧ s^-1. Find how far would the sound travel when the fork just completes 30 vibrations.
Solution:
Here V = 320 m ᐧ s^-1 ; n = 400
We know, V = nλ
∴ 320 = 400 × λ or, λ = \(\frac{320}{400}\) = \(\frac{4}{5}\)m
So, when the fork completes 1 vibrations, sound travels \(\frac{4}{5}\)m.
∴ When the fork completes 30 vibrations, sound travels \(\frac{4}{5}\) × 30 = 24 m.

Example 4.
A light pointer attached to one arm of a tuning fork touches a plate. The turning fork is made to vibrate and the plate is allowed to fall freely downwards simultaneously. The tuning fork completes 8 vibrations when the pointer shows downward displacement of 10 cm of the plate. What is the frequency of the tuning fork?
Solution:
Suppose a time t is taken by the plate to move 10 cm downwards.
So, from the relation h = \(\frac{1}{2}\)gt², we have
10 = \(\frac{1}{2}\) × 980 × t²
or, t² = \(\frac{1}{49}\) or, t = \(\frac{1}{7}\)s
The tuning fork completes 8 vibrations in that time.
So, frequency of the tuning fork = \(\frac{8}{1 / 7}\) = 56 Hz.

Example 5.
What is the length of a compression in the sound wave, produced by a tuning fork of frequency 440 Hz? Given, velocity of sound in air is 330 m ᐧ s^-1.
Solution:
Frequency of the sound wave (n)
= frequency of the tuning fork = 440 s^-1
Velocity of sound (V) = 330 m ᐧ s^-1
∴ Wavelength, λ = \(\frac{V}{n}\) = \(\frac{330}{440}\) = \(\frac{3}{4}\)m
∴ Length of compression = \(\frac{\text { wavelength }}{2}\) = \(\frac{3}{4 \times 2}\)
= 0.375 m = 37.5 cm.

Example 6.
A rod hanging from a spring is dipped partly in water. The rod vibrates 180 times per minute. As a result, waves are formed on water and have 6 consecutive crests within a distance of 30 cm. Find the velocity of the wave in water.
Solution:
Here, frequency n = \(\frac{180}{60}\) = 3 Hz
There are 5 waves within 6 consecutive crests.
So, the length of these 5 waves = 30 cm
∴ Wavelength, λ = \(\frac{30}{5}\) = 6 cm
∴ Velocity of the wave,
V = nλ = 3 × 6 = 18 cm ᐧ s^-1

Example 7.
The frequency of a tuning fork is 280 Hz. Sound wave advance 80 m in a medium while the tuning fork executes 70 complete oscillations. Determine the velocity of sound in that medium. [HS ’06]
Solution:
Wavelength, λ = \(\frac{80}{70}\) = \(\frac{8}{7}\)m
So, the velocity of sound in the medium,
V = nλ = 280 × \(\frac{8}{7}\) = 320 m ᐧ s^-1.

Example 8.
The frequency of a tuning fork is 512 Hz. When the tuning fork makes 30 vibrations, the emitted sound travels 20 m in air. Determine the wavelength and the velocity of sound wave in air.
Solution:
Wavelength, λ = \(\frac{20}{30}\) = 0.667
∴ Velocity of sound in air,
V = nλ = 512 × \(\frac{20}{30}\) = 341.33 m ᐧ s^-1

Example 9.
When two vibrating turning forks of frequencies 50 Hz and 100 Hz touch the surface of water, they produce waves of wavelengths 0.6 cm and 0.36 cm, respectively. Compare the velocities of the two surface waves.
Solution:
If V₁ and V₂ are the velocities of the two surface waves, then
\(\frac{V_1}{V_2}\) = \(\frac{n_1 \lambda_1}{n_2 \lambda_2}\) = \(\frac{50 \times 0.6}{100 \times 0.36}\) = \(\frac{5}{6}\)
i.e., V₁ : V₂ = 5 : 6