Science > Physics > Wave Motion > Numerical Problems on Wave Motion Set – 01
In this article, we shall study to solve problems to calculate frequency, wavelength, and velocity of wave.
Formulae:
v = n λ
n = 1/T
Wave number = 1/λ
Velocity of radio waves = 3 x 108 m s-1
Where,
v = Velocity of wave
n = Frequency of wave
λ = Wavelength of wave
T = Period of wave
Example 01:
A transverse wave has wavelength of 8.0 meters and frequency 200 Hertz. What is the velocity of the wave?
Given: Frequency, n = 200 Hz, Wavelength = λ = 8.0 m.
To Find: Velocity of wave = v =?
Solution:
Velocity of wave is given by
v = nλ = 200 x 8 = 1600 m s-1
Ans: The velocity of wave is 1600 m s-1.
Example 02:
A tuning fork vibrates with a frequency of 356 Hz to produce sound waves which travel through air with a velocity of 330 m/s. Find the wavelength of sound waves produced.
Given: Frequency, n = 356 Hz, Velocity of sound, v = 330 m s-1.
To Find: Wavelength of wave = λ=?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 330/356 = 0.93 m
Ans: Wavelength of wave is 0.93 m.
Example 03:
A radio station broadcasts at a frequency of 12 MHz. Find the wavelength of radio waves.
Given: Frequency, n = 12 MHz = 12 x 106 Hz, Velocity of radio wave = v = 3 x 108 m s-1.
To Find: Wavelength of wave = λ =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 3x 108/12 x 106 = 25 m
Ans: Wavelength of the radio wave is 25 m.
Example 04:
A tuning fork makes one complete vibration in 1/200 second and the velocity of sound waves is 340 m/s. Find the wavelength of the sound given out by the tuning fork.
Given: velocity of sound, v = 340 m/s, Time period = T = 1/200 s.
To Find: Wavelength of wave =?
Solution:
Frequency = 1/T = 1/(1/200) = 200 Hz
Velocity of wave is given by
v = nλ
∴ λ = v/n = 340/200 = 1.7 m
Ans: Wavelength of the wave is 1.7 m.
Example 05:
If a tuning fork vibrates with a frequency of 512 Hz to produce sound waves which travel with a velocity of 330 m s−1. What is the wavelength of the wave?
Given: velocity of sound, v = 330 m/s, Frequency = n = 512 Hz.
To Find: Wavelength of wave = λ =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 330/512 = 0.645 m
Ans: Wavelength of the wave is 0.645 m.
Example 06:
The velocity of radio waves is 3 × 108 m s-1. Find the wavelength of radio waves being broadcast at a frequency of 500 KHz.
Given: Frequency, n = 500 kHz = 500 x 103 Hz, Velocity of radio wave = v = 3 x 108 m s-1.
To Find: Wavelength of wave = λ =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 3x 108/500 x 103 = 600 m
Ans: Wavelength of the radio wave is 600 m.
Example 07:
The speed of wave in a medium is 960 m s-1. If 3600 waves are passing through medium in 1 minute, calculate the wavelength of the wave.
Given: Frequency = n = 3600 waves per minute = 3600/60 = 60 Hz, Velocity of radio wave = v = 960 m s-1.
To Find: Wavelength of wave = λ =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 960/60 = 16 m
Ans: Wavelength of the radio wave is 16 m.
Example 08:
If the frequency of tuning fork is 400Hz and the velocity of sound in air is 320 m/s, find how far the sound travels while the tuning fork makes 30 vibrations.
Given: Frequency, n = 400Hz, Velocity of sound, v = 320 m s-1, No. of vibration made = 30
To Find: Distance travelled by a wave =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 320/400 = 0.8 m
Distance travelled by sound = Wavelength x No. of Vibrations = 0.8 X 30
∴ Distance travelled by sound = 24.0 m
Ans: Distance travelled by the sound wave in 30 vibrations is 24 m.
Example 09:
If the frequency of tuning fork is 256Hz and the velocity of sound in air is 320 m/s, find how far the sound travels while the tuning fork makes 64 vibrations.
Given: Frequency, n = 256Hz, Velocity of sound, v = 320 m s-1, No. of vibration made = 64
To Find: Distance travelled by a wave =?
Solution:
Velocity of wave is given by
v = nλ
∴ λ = v/n = 320/256 = 1.25 m
Distance travelled by sound = Wavelength x No. of Vibrations = 1.25 X 64
∴ Distance travelled by sound = 80 m
Ans: Distance travelled by the sound wave in 64 vibrations is 80 m.
Example 10:
Audible frequencies have a range 20 Hz to 20 kHz. Express this range in terms of (i) period T (ii) wavelength λ in air and (iii) angular frequency. Given velocity of sound in air 330 m s-1.
Given: Frequency range n1 = 20 Hz to n2 = 20 kHz = 20 x 103 Hz, Velocity of sound in air = v = 330 ms-1.
To Find: Time period range T1 to T2, Wavelength range λ1 to λ2, Angular frequency range ω1 to ω2.
Solution:
Audible range in terms of period T:
T1 = 1/n1 = 1/20 = 5 x 10-2 s
T2 = 1/n2 = 1/20 x 103 = 5 x 10-5 s
Thus, the audible range in terms of period is from 5 x 10-2 s to 5 x 10-5 s.
Audible range in terms of wavelength λ:
Velocity of wave is given by
v = nλ
∴ λ1 = v/n1 = 330/20 = 16.5 m
∴ λ2 = v/n2 = 330/20 x 103 = 0.0165 m
Thus, the audible range in terms of wavelength is from 16.5 m to 0.0165 m.
Audible range in terms of angular frequency ω:
We have ω = 2πn
∴ ω1 = 2πn1 = 2π x 20 = 40π rad s-1
∴ ω2 = 2πn2 = 2π x 20 x 103 = 40000π rad s-1
Thus, the audible range in terms of angular frequency is from 40π rad s-1 to 40000π rad s-1.
Example 11:
A tuning fork produces sound waves of wavelength 68 cm. If the velocity of sound is 340 m/s. What is the frequency of tuning fork?
Given: Velocity of wave = v = 340 m s-1, Wavelength = λ = 68 cm = 68 x 10-2 m.
To Find: Frequency = n =?
Solution:
Velocity of wave is given by
v = nλ
∴ n = v/λ = 340/68 x 10-2 = 500 Hz
Ans: Frequency of tuning fork is 500 Hz.
Example 12:
A radio station broadcasts a programme at 219.3 m. Determine frequency of radio waves.
Given: Wavelength of wave = λ = 219.3 m, Velocity of radio wave = v = 3 x 108 m s-1.
To Find:?
Solution:
Velocity of wave is given by
v = nλ
∴ n= v/ λ = 3x 108/219.3 x 103
∴ n = 1.368 x 106 Hz = 1.368 MHz
Ans: Frequency of the radio wave is 1.368 MHz
Example 13:
A wave train moving along a rope has a velocity of 100 m s-1 and wavelength of 20 m. What is its time period?
Given: Velocity of wave = v = 100 m s-1, Wavelength = λ = 20 m.
To Find: Period = T =?
Solution:
Velocity of wave is given by
v = nλ
∴ n = v/λ = 100/20 = 5 Hz
Period = T = 1/n = 1/5 = 0.2 s
Ans: Period of wave is 0.2 s.