CG Board Class 9 Science Chapter 14 Solutions: Sound (ध्वनि) – 2026-27

(i) When sound waves propagate through a medium, then from one position to another:**

• (a) Particles of the medium travel
• (b) Transfer of energy from one particle to another takes place
• (c) Transformation of energy takes place
• (d) None of these

(ii) Sound below the frequency of 20 Hz is called:

• (a) amplitude
• (b) inaudible/infrasound
• (c) ultrasound
• (d) none of these

(iii) The distance between consecutive compressions or rarefactions is called:

• (a) amplitude
• (b) frequency
• (c) speed of wave
• (d) wavelength

(iv) Establish a relation between velocity of transferred wave (v), its frequency (ν) and wavelength (λ):

• (a) v = (ν)/(λ)
• (b) v = ν λ
• (c) v = (λ)/(ν)
• (d) none of these

(v) Sound waves do not travel in:

• (a) solids
• (b) liquids
• (c) gases
• (d) vacuum

(vi) If the time period of a vibrating object is 0.05 s, the frequency of waves produced will be:

• (a) 5 Hz
• (b) 20 Hz
• (c) 200 Hz
• (d) 2 Hz

(i) The hearing range of sound in humans is …………………..**

(ii) SI unit of wavelength is …………………………..

(iii) Distance between consecutive compression or consecutive rarefactions is called ……………….

(iv) Velocity of sound depends on ……………………………..

• The velocity of sound in a given medium remains unchanged (constant).
• Reason: The speed of sound depends only on the nature, temperature, density, and elasticity of the medium, not on the frequency or wavelength of the sound source. If frequency (ν) is doubled, the wavelength (λ) becomes half, keeping their product v = ν λ constant.

• English Step-by-Step:
• Let frequency of source B = ν_B. Then frequency of source A = ν_A = 2ν_B.
• Since both sounds travel in the same medium, their speed v is the same.
• Formula: v = ν_A λ_A = ν_B λ_B
• Substitute ν_A = 2ν_B: 2ν_B × λ_A = ν_B × λ_B ⇒ 2 λ_A = λ_B ⇒ λ_A = (λ_B)/(2)
• Conclusion: The wavelength of sound A is half the wavelength of sound B (Ratio λ_A : λ_B = 1 : 2).
• Hindi — :
• B = ν_B A = ν_A = 2ν_B
• , v
• : v = ν_A λ_A = ν_B λ_B
• ν_A = 2ν_B : 2ν_B × λ_A = ν_B × λ_B ⇒ 2 λ_A = λ_B ⇒ λ_A = (λ_B)/(2)
• : A B **

The *Quality or Timbre of sound helps us identify our friend’s voice. Timbre depends on the shape of the sound wave and the overtone frequencies produced by the unique structure of vocal cords.

Water (liquid medium)** will produce a louder sound.

• Reason: Water has higher density and elasticity than air, so the pressure created by vibrations is higher and energy dissipation during propagation is significantly lesser, resulting in a clearer and louder sound underwater.

Sound:** Sound is a form of energy that produces the sensation of hearing in our ears. It travels as a mechanical longitudinal wave consisting of compressions and rarefactions.

• Production: Sound is produced by the vibrations (rapid back-and-forth motion) of an object. When an object vibrates, it exerts force on surrounding medium particles, creating waves.

Compression (C):** When a vibrating body moves forward, it pushes and compresses the air in front of it, creating a high-pressure and high-density region called compression.

• Rarefaction (R): When the vibrating body moves backwards, it creates a low-pressure and low-density region called rarefaction.
• Diagrammatic representation: Series of dense lines/coils for C and spread out lines/coils for R.

• English Activity (Bell-Jar Experiment):

1. Hang an electric bell inside a glass bell jar connected to a vacuum pump.
2. Press the switch: the bell rings and the sound is clearly heard.
3. Start the vacuum pump to withdraw air gradually from the jar: the sound becomes fainter.
4. When all the air is completely pumped out (vacuum is created), no sound can be heard even though the hammer continues to strike the gong.
5. Conclusion: Sound cannot travel in a vacuum; it strictly requires a material medium.

• Hindi :

1.

2.

1. –
2. ,
3. : ;

A wave in which the particles of the medium vibrate back and forth along a direction parallel to the direction of wave propagation is called a longitudinal wave.

• As sound travels through air or liquid, medium particles vibrate parallel to the direction in which sound energy travels, creating continuous compressions and rarefactions. Hence, sound waves are longitudinal mechanical waves.

(i) Echocardiography:**

A medical diagnostic technique where ultrasonic waves are reflected from various parts of the heart to produce detailed images of heart structure and cardiac movements.

• (ii) Ultrasonography:

A technique using an ultrasound scanner to generate images of internal organs (liver, gall bladder, kidneys, uterus, fetus during pregnancy) by detecting reflected waves from tissue density boundaries.

• (iii) SONAR:

Stands for SOund NAvigation And Ranging. A ship-mounted device containing a transmitter and detector that uses ultrasonic echo signals to measure depth, distance, and speed of underwater objects (d = vt/2).

• Definitions / :
• Wavelength (λ): Distance between two consecutive compressions or crests.
• Frequency (ν): Number of complete oscillations per unit time.
• Velocity (v): Distance covered by a wave per unit time (v = (Distance)/(Time)).
• In time period T, the wave covers a distance equal to its wavelength λ.
• Speed (v) = (Wavelength (λ))/(Time Period (T)) = (λ)/(T)
• Since frequency ν = 1/T
• v = ν λ (or Velocity = Frequency × Wavelength).

• English Step-by-Step:
• Given speed (v) = 3329 m/s
• Wavelength (λ) = 1.5 cm = 1.5/100 m = 0.015 m
• Formula: v = ν λ ⇒ ν = (v)/(λ)
• Calculation: ν = (3329)/(0.015) = 221,933.33 Hz (or ≈ 221.93 kHz)
• (Note: If text speed is read as 332 m/s: ν = (332)/(0.015) = 22,133.33 Hz).
• Audibility: Human audible range is 20 Hz to 20,000 Hz (20 kHz). Since 221.93 kHz > 20 kHz, this wave is ultrasonic and NOT audible to human ears.
• Hindi — :
• : (v) = 3329 m/s
• (λ) = 1.5 cm = 0.015 m
• : ν = (v)/(λ)
• : ν = (3329)/(0.015) = 221,933.33 Hz
• : 20 Hz 20,000 Hz 20 kHz ,

• English Step-by-Step:

1. Find Frequency of Source in Air:

• v_air = 340 m/s, λ_air = 0.017 m
• ν = (v_air)/(λ_air) = (340)/(0.017) = 20,000 Hz (20 kHz)

1. Frequency Remains Constant across media:

• Frequency depends solely on the source, so ν = 20,000 Hz in water as well.

1. Calculate New Wavelength in Water (λ_water):

• v_water = 1480 m/s
• λ_water = (v_water)/(ν) = (1480)/(20000) = 0.074 m (or 7.4 cm)

1. Effect: The wavelength increases from 0.017 m to 0.074 m (increases by a factor of (1480)/(340) ≈ 4.35).

• Hindi — :

1. :

• v = 340 m/s, λ = 0.017 m
• ν = (340)/(0.017) = 20,000 Hz

1. :

• ν = 20,000 Hz

1. :

• v = 1480 m/s
• λ = (1480)/(20000) = 0.074 m (7.4 cm)

1. : 0.017 m 0.074 m

Pitch depends directly on frequency. To reduce pitch (frequency), we can

1. Decrease Tension: Loosen the tuning peg to decrease tension in the sitar string.
2. Increase Vibrating Length: Press the string at a fret that increases the active vibrating length of the string.
3. Use Thicker String: Pluck a thicker string with greater mass per unit length.

• English Step-by-Step:
• Wavelength (λ) = 1 cm = 1/100 m = 0.01 m
• Velocity (v) = 330 m/s
• Formula: ν = (v)/(λ)
• Calculation: ν = (330)/(0.01) = 33,000 Hz (or 33 kHz)
• Audibility: Human hearing range is 20 Hz to 20,000 Hz (20 kHz). Since 33,000 Hz > 20,000 Hz, this is an ultrasonic wave and CANNOT be heard by a normal human being.
• Hindi — :
• (λ) = 1 cm = 0.01 m
• (v) = 330 m/s
• : ν = (v)/(λ)
• : ν = (330)/(0.01) = 33,000 Hz
• : 20 Hz 20,000 Hz 33,000 Hz (20 kHz) ,