Wavelength of a wave decreases when frequency increases and wavelength of a wave increases when frequency decreases. As the frequency increases, the distance between consecutive wave crests or troughs decreases. As the frequency decreases, the distance between consecutive wave crests or troughs increases. Wavelength of a wave is inversely proportional to the frequency of wave.
Frequency and Wavelength
In physics, frequency and wavelength are important characteristics related to a wave cycle. Now, what is a wave? The transfer of energy from one point to another by a disturbance or variation is known as a wave. Now, the wavelength is defined as the distance between two successive points in phase with each other, whereas frequency is defined as the number of waves produced per second.
Frequency
The total number of wave cycles or oscillations produced per unit time is called the frequency (f) of a wave. It is measured in terms of Hertz (Hz) or s-1. For humans, the audible range of sound frequencies is from 20 Hz to 20 kHz. Human ears cannot hear ultrasonic sounds, i.e., the frequencies above the audible range, and also infrasound, i.e., the frequencies less than the audible range.
The formula for the frequency of a wave is,
Frequency (f) = 1/T
Where T is the period of a wave.
- The period of a wave is defined as the time taken by a wave to complete one full cycle or to complete an oscillation.
- From the formula of frequency, we can observe that the frequency of a wave is inversely proportional to its period.
1 Hertz = 1 oscillation/second
Wavelength
The length of the distance between the two successive crests or troughs of a wave is called the wavelength. The crest is the highest point of the wave, whereas the lowest point of the wave is the trough. As the wavelength is a distance or length between two points, it is measured in meters, centimeters, millimeters, micrometers, Angstroms, etc. It is denoted by the Greek symbol Lambda ‘λ’.
The formula for the wavelength is,
Wavelength(λ) = velocity(v)/frequency(f)
⇒ λ = v/f
- The distance traveled by the wave in a unit of time is called the velocity of a wave or wave velocity. The S.I. unit of wave velocity is ms-1.
- The velocity of a wave is equal to the product of its wavelength and frequency.
v = λ × f
The distance travelled by a wave in a unit of time is equal to one wavelength.
⇒ Wave velocity (v) = Wavelength(λ)/Period(T)
We know that,
frequency (f) = 1/Period(T)
⇒ Wave velocity (v) = Wavelength(λ) × Frequency(f)
⇒ v = λ × f
⇒ λ = v/f
How does the wavelength of a wave change when frequency decreases when frequency increases?
Answer:
Let’s consider a thread that is tied to an end. Now hold the other end of the rope and oscillate it faster, resulting in higher frequency waves. We can also observe that the waves are produced with a shorter wavelength. Hence, we can draw the conclusion that there is a relationship between wavelength and frequency.
From the equation of the wavelength, we can tell that the wavelength of a wave is inversely proportional to its frequency, i.e., as the frequency of a wave increases, its wavelength decreases. Similarly, as the frequency of a wave decreases, its wavelength increases.
λ ∝ 1/f
For example, let’s consider a wave whose new frequency is two times its old frequency. Now, what will be the new wavelength of the wave?
Given data,
Let λ and λ’ be the old and new wavelengths of the light wave.
Let f be the old frequency of the wave.
Now, the new frequency of the wave = 2f
Wavelength (λ) = Velocity/Frequency
The velocity of the wave remains constant.
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ λ/λ’ = (2f)/f = 2
⇒ λ’ = λ/2
Hence, the new wavelength of the wave is half of the old wavelength.
Therefore, as the frequency of a wave increased, its wavelength decreased.
Therefore, as the frequency of a wave decreased, its wavelength increased.
- A wave with a high frequency has a short wavelength and high energy.
- A wave with a low frequency has a long wavelength and low energy.
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Sample Problems
Problem 1: Two sound waves are traveling through the air at the same speed. If a wave whose frequency is 45 kHz has a wavelength of 7.5 mm, then find the frequency of the wave whose wavelength is 100 nm.
Solution:
Given data,
The speed of both the waves is equal ⇒ v1 = v2 = v, where v is the velocity of a sound wave
The wavelength of the first wave ( λ1) =7.5 mm = 7.5 × 10-3 m
Frequency of the first wave (f1) = 45 kHz = 45 × 103 Hz
Wavelength of second wave (λ2) = 100 nm = 100 × 10-9 m
Frequency of the second wave (f2) =?
Wavelength (λ) = Velocity/Frequency
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ (7.5 × 10-3)/(100 × 10-9) = f2/(45 × 103)
⇒ f2 = [(7.5 × 10-3) × (45 × 103)]/(100 × 10-9)
⇒ f2 = 33.75 GHz
Hence, the frequency of the wavelength of 100 nm is 33.75 GHz.
Therefore, as the wavelength of a wave decreased, its frequency increased.
Problem 2: A light wave is traveling in a vacuum. What will be the new wavelength of the wave if the new frequency of the wave is one-fifth of its old frequency?
Solution:
Given data,
Let λ and λ’ be the old and new wavelengths of the light wave.
Let f be the old frequency of the wave.
Now, the new frequency of the wave = f/5
Wavelength (λ) = Velocity/Frequency
The velocity of the wave remains constant.
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ λ/λ’ = (f/5)/f = 1/5
⇒ λ’ = 5λ
Hence, the new wavelength of the wave is five times the old wavelength.
Therefore, as the frequency of a wave decreased, its wavelength increased.
Problem 3: Two sound waves traveling through water at the same speed. If a wave whose frequency is 3 MHz has a wavelength of 100 m, then find the wavelength of the wave whose frequency is 4.5 kHz.
Solution:
Given data,
The wavelength of the first wave ( λ1) = 100 m
Frequency of the first wave (f1) = 3 MHz = 3 × 106 Hz
Frequency of the second wave (f2) = 4.5 kHz = 4.5 × 103 Hz
The wavelength of the second wave ( λ2) = ?
Speed of both the waves is equal ⇒ v1 = v2 = v
Wavelength (λ) = Velocity/Frequency
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ 100/ λ2 = (4.5 × 103)/(3 × 106)
⇒ λ2 = (3 × 106 × 100 )/(4.5 × 103) = 66.67 km
Hence, the wave of frequency 4.5 kHz wavelength is 66.67 km.
Therefore, as the frequency of a wave decreased, its wavelength increased.
Problem 4: A light wave is traveling in a vacuum. What will be the new wavelength of the wave if the new frequency of the wave is three times its old frequency?
Solution:
Given data,
Let λ and λ’ be the old and new wavelengths of the light wave.
Let f be the old frequency of the wave.
Now, the new frequency of the wave = 3f
Wavelength (λ) = Velocity/Frequency
The velocity of the wave remains constant.
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ λ/λ’ = 3f/f = 3
⇒ λ’ = λ/3
Hence, the new wavelength of the wave is one-third of the old wavelength.
Therefore, as the frequency of a wave increased, its wavelength decreased.
Problem 5: A light wave is traveling in a vacuum. What will be the new frequency of the wave if the new wavelength of the wave is four times its old wavelength?
Solution:
Given data,
Let f and F be the old and new frequencies of the light wave.
Let λ be the old wavelength of the wave.
Now, the new wavelength of the wave = 4λ
Wavelength (λ) = Velocity/Frequency
The velocity of the wave remains constant.
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ λ /4λ = F/f
⇒ F/f = 1/4 ⇒ F = f/4
Hence, the new frequency of the wave is one-fourth of the old frequency.
Therefore, as the wavelength of a wave increased, its frequency decreased.
Problem 6: Two sound notes produced by a tuning fork are traveling at the same speed. If a wave whose frequency is 6 kHz has a wavelength of 100 mm, then find the wavelength of the wave whose frequency is 5.4 kHz.
Given data,
The wavelength of the first wave ( λ1) = 100 × 10-3 m = 0.1 m
Frequency of the first wave (f1) = 6 kHz = 6 × 103 Hz
Frequency of the second wave (f2) = 5.4 kHz = 5.4 × 103 Hz
The wavelength of the second wave ( λ2) =?
Speed of both the waves is equal ⇒ v1 = v2 = v
Wavelength (λ) = Velocity/Frequency
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ (0.1)/ λ2 = (5.4 × 103)/(6 × 103)
⇒ λ2 = (6 × 103 × 0.1 )/(5.4 × 103) = 0.111 m = 111 mm
Hence, the wave of frequency 4.5 kHz wavelength is 111 mm.
Therefore, as the frequency of a wave decreased, its wavelength increased.
Problem 7: If the new frequency of an electromagnetic wave is two-thirds of its old frequency, then what will be the new wavelength of the wave?
Solution:
Given data,
Let λ and λ’ be the old and new wavelengths of the light wave.
Let f be the old frequency of the wave.
Now, the new frequency of the wave = (2/3)f
Wavelength (λ) = Velocity/Frequency
The velocity of the wave remains constant.
⇒ λ ∝ 1/f
⇒ λ1/ λ2 = f2/ f1
⇒ λ/λ’ = (2/3)f/f = 2/3
⇒ λ’ = 3λ/2 = 1.5λ
Hence, the new wavelength of the wave is three and a half times its old wavelength.
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