Relation Between Frequency and Velocity

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Frequency and Velocity are the two fundamental characteristics of a wave. Waves refer to a propagating disturbance that travels through a medium transferring energy from one place to another. There is no net movement of particles within the medium.

  • Frequency is the number of waves passing through one point in per unit of time.
  • Wave Velocity is defined as the distance traveled by a wave per unit of time. 
  • Frequency is denoted by the letter ‘f’ and velocity is represented by ‘V’.
  • The SI unit of frequency and wave velocity are Hz and m/s respectively. 

Frequency and Velocity are directly proportional to each other. An increase in frequency will increase the velocity of the wave for a constant wavelength. Relation between Frequency and Wavelength is expressed as follows

\(V=f\times \lambda\)

Here, V is the wave velocity, f is the frequency and λ is the wavelength of the wave. 

Read More: NCERT Solutions for Class 11 Physics Waves

Key Terms: Frequency, Wave Velocity, Velocity, Wavelength, Crest, Trough, Hertz, Phase Velocity, Waves, Wave Velocity Formula


What is Frequency?

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Frequency of a wave is the number of waves produced in a given interval of time. Frequency is measured by counting the number of crests or troughs that pass a point in one second.

  • It is the number of cycles or turns made by a wave per second.
  • The unit of frequency is Hertz (Hz)
  • Waves that are moving faster will have a higher frequency than waves that are moving slowly.
  • Frequency is calculated by dividing the wave speed (v) by the wavelength (λ). 
  • For example, if a wave completes three cycles in 1 second, the frequency of the wave is 3 Hertz.
Frequency
Frequency

Frequency Formula is given as  

\(f={v\over \lambda}\)

Read More: Waves Important Questions 

Solved Example

Example: What will be the frequency of a wave that is traveling at a speed of 3 m/s and has a 2-meter wavelength?

Solution: Given that

  • Speed of Wave (v) = 3 m/s
  • Wavelength (λ) = 2 m

Using the Frequency Formula, we get

\(f={v\over \lambda}\)

f = 3/2 = 1.5 Hz

Thus, the frequency of the given wave is 1.5 Hz.


What is Wave Velocity?

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Wave velocity is the distance traveled by a wave in a given amount of time.

  • It is the speed at which a disturbance or a wave propagates in a given medium.
  • It is determined by the nature of the medium in which the wave is traveling. 
  • Wave velocity is also referred to as ‘Phase Velocity’.
  • It is denoted by the letter ‘v’ and its SI unit is m/s.
  • Wave velocity is the product of the wavelength and frequency of the wave.

Wave Velocity Formula is given as 

\(v=f\times \lambda\)

Read More: Wavelength Formula

Solved Example

Example: Calculate the velocity of a wave with a 10 m wavelength and a 16 Hz frequency.

Solution: Given that

  • Frequency of Wave (f) = 16 Hz
  • Wavelength of Wave (λ) = 10 m

Using the Wave Velocity Formula, we get

V = λf

V = 16 × 10 = 160 m/s

Thus, the velocity of the given wave is 160 m/s.


Relation between Frequency and Velocity

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Frequency is the number of waves per second while Wave Velocity is the distance traveled by a wave per unit of time. For a particular wave, Wave velocity is given as the product of the wavelength and frequency.

Thus, the Relation between Frequency and Velocity is given as

\(V=f\times \lambda\)

Where

  • V: Velocity of the Wave in m/s.
  • f: Frequency of the Wave in Hz.
  • λ: Wavelength of the wave in m.

From the given wave equation, it is derived that,

“Wave Velocity is directly proportional to the Frequency of the wave.”

Thus, it can be concluded that if the frequency of a wave increases, the velocity of the wave will also increase. For instance, 

  • If the frequency is doubled, the velocity of the wave will also be doubled.
  • If the frequency is increased four times, the velocity of the wave will also increase by four times.

Relation between frequency and velocity holds true for all types of waves such as water waves, sound waves, and visible light

Parts of a Wave

Parts of a Wave

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Things to Remember

  • Frequency of waves is the number of waves that pass through a fixed point in per unit of time.
  • It is denoted by the letter ‘f’ and its SI unit is Hertz (Hz).
  • Velocity of a wave is the speed at which a wave propagates in a medium.
  • The SI unit of Wave velocity is m/s and is denoted by the letter ‘v’.
  • The velocity of a wave is constant in a given material.
  • Relation between Frequency and Velocity is \(V=f\times \lambda\).
  • It states that the frequency and velocity of a wave are directly proportional to each other.

Read More: Waves MCQs


Previous Years’ Questions

  1. The velocity of sound waves in a medium does not depend on… (UPSEE 2019)
  2. The velocity of the electromagnetic wave is parallel to… (NEET 2002)
  3. A wave has velocity v in medium P and velocity 2v in medium… (BITSAT 2007)
  4. Velocity of sound waves in air is 330m/s. For a particular sound…
  5. The velocity of all radio waves in free space is… (Rajasthan PMT 2005)
  6. An electromagnetic wave of frequency υ = 3.0 MHz… (NEET 2013)
  7. A wave of frequency 500 Hz travels with a speed of 360 m/s… (KEAM)
  8. A wave of frequency 100 Hz travels along a string towards its fixed… (NEET 1994)
  9. A traveling acoustic wave of frequency 500 Hz is moving along… (WBJEE 2013)
  10. Three sound waves of equal amplitudes have frequencies… (NEET 2016)

Sample Questions

Ques. A man creates two waves every second by shaking the spring up and down. If each wave travels 0.9 m after one complete wave cycle, what is the velocity of wave propagation? (3 Marks)

Ans. Wavelength is the distance a wave has traveled after one complete cycle or one period. It is given that,

  • Wavelength (λ) = 0.9 m
  • Frequency (f) = 2 Hz
  • Velocity v =?

Using the Relation between Velocity and Frequency, we get

\(V=f\times \lambda\)

V =  2 x (0.9 m) = 1.8 m/s

Thus, the velocity of wave propagation is 1.8m/s.

Ques. Calculate the wave velocity of the ocean wave if the distance between wave crests is 10 m and the time for a ball floating on the surface to bob up and down is 5 sec. (3 Marks)

Ans. It is given that,

  • Wavelength λ = 10.0 m
  • Time T = 5.00 s
  • Wave Velocity V =?

Using the equation V = λ/T to find the wave velocity, we get

V = 10m / 5 s = 2.00 m/s

Thus, the wave velocity of the ocean wave is 2m/s.

Ques. What are the types of waves? (3 Marks)

Ans. Waves can be classified into two major types: 

  • Transverse Waves: Transverse wave is a wave, in which individual particles of the medium move up and down in a direction perpendicular to the direction of wave propagation. These waveforms crest and trough. Ripples in water are examples of transverse waves.
  • Longitudinal Waves: Longitudinal wave is a wave in which individual particles move to and fro in the same direction along which the wave propagates. This waveforms compression and rarefaction. Sound waves are a good example of longitudinal waves. 

Ques. A fork vibrates with a frequency of 256. If the velocity of sound is 345.6 m/s. Find the wavelength and distance, which the sound travels during the time the fork makes 50 vibrations. (3 Marks)

Ans. According to the question, 

  • Frequency f = 256
  • Wave Velocity V = 345.6 m/s
  • Wavelength λ =?

Using the Relation between Velocity and Frequency, we get

\(V=f\times \lambda\)

λ = V/f

λ = 345.6/256 = 1.35m

Thus, the wavelength is 1.35 m.

The distance traveled in 50 vibrations = 50 × 1.35 = 67.5m

Ques. If the frequency of a tuning fork is 400, find the distance the sound waves travel when the tuning fork makes 30 vibrations. The velocity of sound in air is 320 m/sec. (3 Marks)

Ans. Given that,

  • Frequency f = 400 Hz
  • Velocity v = 320 m/s
  • Wavelength λ =?

Using the Relation between Velocity and Frequency, we get

\(V=f\times \lambda\)

λ = V/f

λ = 320/400 m

  • Distance travelled in 1 vibration = λ= 320/400m
  • Distance traveled in 30 vibrations= (320/400) 30 = 24m.

Thus, the distance traveled by the wave is 30 vibrations is 24 m.

Ques. From a radio station, the frequency of radio waves is 15 megacycles/sec. Calculate the wavelength. (3 Marks)

Ans. It is given that,

Frequency f = 15 megacycle/sec = 15 x106 cycles/sec

Since the velocity of radio waves is equal to the velocity of light in a vacuum, thus, 

V = 3 × 108m/s

λ =V/f

λ = (3 × 108m/s)/(15 x106) = 20m

Thus, the wavelength of the radio wave is 20 m.

Ques. Audible frequencies range from 40 Hz to 30,000 Hz. What is the range in terms of wavelength in the air? (Velocity of Sound in Air 350 m/s) (3 Marks)

Ans. Given that

  • V = 350 m/s
  • Range = 40 Hz to 30,000 Hz

We need to find the wavelengths of both frequencies, thus, 

λ =V/f

  • For 40 Hz, λ = 350/40 = 8.75 m
  • For 30,000 Hz, λ = 350/ 30,000 = 0.012 m

Thus, the range in terms of wavelength is from 8.75 m to 0.012 m.

Ques. The wavelength of a wave is 6 m and the frequency is 12 Hz. What will be the velocity of the wave? (2 Marks)

Ans. According to the question, 

  • Frequency f = 12 Hz
  • Wavelength λ = 6 m

Using the Wave Velocity Formula, 

V = λν 

V = 12 × 6 = 72 m/s

Thus, the velocity of the wave is 72 m/s.

Ques. What will be the frequency of the pendulum that takes 8 seconds to complete one cycle? (2 Marks)  

Ans. According to the question, time is given as 8 seconds. 

Using the Frequency Formula, 

f = 1 / T

f = 1 / 8 = 0.125

Thus, the frequency of the pendulum is 0.125 Hz.

Ques. Determine the wavelength of a wave with a frequency of 120 Hz traveling at a speed of 287 m/s. (3 Marks)

Ans. According to the question,

  • Velocity V = 287 m/s
  • Frequency f = 120 Hz

Using the Relation between Velocity and Frequency, we get

\(V=f\times \lambda\)

λ = V/ν

λ = 287 / 120 = 2.39 m

Thus, the wavelength of the wave is 2.39 m.


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CBSE CLASS XII Related Questions

  • 1.
    Figure shows variation of Coulomb force (F) acting between two point charges with \( \frac{1}{r^2} \), \( r \) being the separation between the two charges \( (q_1, q_2) \) and \( (q_2, q_3) \). If \( q_2 \) is positive and least in magnitude, then the magnitudes of \( q_1, q_2 \), and \( q_3 \) are such that:
    variation of Coulomb force

      • \( q_2<q_1<q_3 \)
      • \( q_3<q_1<q_2 \)
      • \( q_1<q_2<q_3 \)
      • \( q_2<q_3<q_1 \)

    • 2.
      The electric field at a point in a region is given by \( \vec{E} = \alpha \frac{\hat{r}}{r^3} \), where \( \alpha \) is a constant and \( r \) is the distance of the point from the origin. The magnitude of potential of the point is:

        • \( \frac{\alpha}{r} \)
        • \( \frac{\alpha r^2}{2} \)
        • \( \frac{\alpha}{2r^2} \)
        • \( -\frac{\alpha}{r} \)

      • 3.
        A parallel plate capacitor has plate area \( A \) and plate separation \( d \). Half of the space between the plates is filled with a material of dielectric constant \( K \) in two ways as shown in the figure. Find the values of the capacitance of the capacitors in the two cases. parallel plate capacitor


          • 4.
            A charge \( -6 \mu C \) is placed at the center B of a semicircle of radius 5 cm, as shown in the figure. An equal and opposite charge is placed at point D at a distance of 10 cm from B. A charge \( +5 \mu C \) is moved from point ‘C’ to point ‘A’ along the circumference. Calculate the work done on the charge.
            work done on the charge


              • 5.
                A rectangular glass slab ABCD (refractive index 1.5) is surrounded by a transparent liquid (refractive index 1.25) as shown in the figure. A ray of light is incident on face AB at an angle \(i\) such that it is refracted out grazing the face AD. Find the value of angle \(i\).
                A rectangular glass slab ABCD (refractive index 1.5)


                  • 6.
                    Answer the following giving reason:
                    (a) All the photoelectrons do not eject with the same kinetic energy when monochromatic light is incident on a metal surface.
                    (b) The saturation current in case (a) is different for different intensity.
                    (c) If one goes on increasing the wavelength of light incident on a metal sur face, keeping its intensity constant, emission of photoelectrons stops at a certain wavelength for this metal.

                      CBSE CLASS XII Previous Year Papers

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