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

A person standing outside cannot hear clearly the sound of a person sitting inside a closed car, a closed bus or a closed room. But when you knock on the door or window of a car, the sound can be heard on both the sides. Why is it so?

• Air inside a closed car is isolated from the outside air by solid glass/metal walls. When a person speaks inside, most of the sound waves in air get reflected back from the solid glass/metal boundary because of the huge difference in medium density and elasticity between air and solid.
• However, when you knock directly on the glass window or door, your action vibrates the solid material directly. Since solids are dense and highly elastic, the vibrations travel efficiently through the solid and vibrate the air particles on both the inner and outer sides of the door/window, making the sound audible on both sides.

Does the length of scale left free affect the sound produced? Explain.

Yes, the free length of the vibrating scale directly affects the pitch and frequency of the sound produced. A shorter free length vibrates faster (higher frequency), producing a sharper or high-pitched sound, whereas a longer free length vibrates slower (lower frequency), producing a deep or low-pitched sound.

The sound of a motor boat can be heard louder under the water as compared to above the surface of water. Why?

Transmission of sound depends on the density and elasticity of the medium. Water (a liquid) has a higher particle density and greater elasticity than air (a gas). Consequently, sound energy travels faster with significantly lesser energy loss/dissipation in water than in air, making the motor boat sound much louder underwater.

What effect will be produced on sound if we move upwards above the surface of Earth?

As we go higher above the Earth’s surface, the atmospheric pressure and air density decrease. As air particles become rarer, sound transmission becomes weaker and slower, and the loudness of sound decreases. At extreme altitudes where air is absent (vacuum), sound cannot travel at all.

Can you hear sound in space? Give reasons.

No, sound cannot be heard in outer space. Sound is a mechanical wave that requires a material medium (particles) for the propagation of compressions and rarefactions. Since outer space is a vacuum with no air or material medium, sound waves cannot propagate.

On a rainy day, we hear the sound of thunder after we see the lightning, why is it so?

Light travels at an extremely high speed (3 × 10⁸ m/s) compared to the speed of sound in air (about 340 m/s). Though both lightning and thunder occur simultaneously in clouds, the light flash reaches our eyes almost instantaneously, whereas the sound of thunder takes significantly more time to cover the same distance through air.

If there is no atmosphere on the moon, will astronauts be able to speak to each other directly?

No, astronauts cannot hear each other directly on the Moon because sound requires a material medium to propagate, and the Moon lacks an atmosphere. They communicate with each other using radio waves built into their helmets, as radio waves are electromagnetic waves that can travel through vacuum.

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

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

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

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

(v) Sound waves do not travel in:

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

(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 ……………………………..

• 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 पद-दर-पद हल:
• माना स्रोत B की आवृत्ति = ν_B। अतः स्रोत A की आवृत्ति = ν_A = 2ν_B।
• चूंकि दोनों ध्वनियां एक ही माध्यम में संचरित होती हैं, अतः उनकी चाल v समान होगी।
• सूत्र: v = ν_A λ_A = ν_B λ_B
• ν_A = 2ν_B रखने पर: 2ν_B × λ_A = ν_B × λ_B ⇒ 2 λ_A = λ_B ⇒ λ_A = (λ_B)/(2)
• निष्कर्ष: ध्वनि A का तरंगदैर्ध्य ध्वनि B के तरंगदैर्ध्य का आधा होगा (अनुपात λ_A : λ_B = 1 : 2)।

The गुण / स्वरता 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 गतिविधि (बेल-जार प्रयोग):

1. कांच के बेलजार के अंदर एक विद्युत घंटी लटकाकर उसे वैक्यूम (निर्वात) पंप से जोड़ते हैं।
2. स्विच दबाने पर घंटी बजती है और उसकी आवाज़ स्पष्ट सुनाई देती है।
3. निर्वात पंप चलाकर धीरे-धीरे बेलजार से हवा बाहर निकालते हैं: घंटी की आवाज़ धीमी होती जाती है।
4. जब पूरी हवा बाहर निकल जाती है (निर्वात बन जाता है), तो घंटी का हथौड़ा टकराते दिखने पर भी कोई आवाज़ सुनाई नहीं देती।
5. निष्कर्ष: ध्वनि निर्वात में संचरित नहीं हो सकती; इसके संचरण हेतु माध्यम अनिवार्य है।

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 (इकोकार्डियोग्राफी):**

• (ii) Ultrasonography (अल्ट्रासोनोग्राफी):
• (iii) SONAR (सोनार):

• Definitions / परिभाषाएं:
• 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 पद-दर-पद हल:
• दिया गया है: चाल (v) = 3329 m/s
• तरंगदैर्ध्य (λ) = 1.5 cm = 0.015 m
• सूत्र: ν = (v)/(λ)
• गणना: ν = (3329)/(0.015) = 221,933.33 या ≈ 221.93 kHz
• श्रव्यता: मानव कर्ण की श्रव्य सीमा 20 Hz से 20,000 Hz है। चूंकि यह आवृत्ति 20 kHz से अधिक है, अतः यह एक पराश्रव्य (अल्ट्रासोनिक) तरंग है और मानव कानों द्वारा श्रव्य नहीं होगी।

• 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 पद-दर-पद हल:

1. स्रोत की आवृत्ति ज्ञात करना:

• v_वायु = 340 m/s, λ_वायु = 0.017 m
• ν = (340)/(0.017) = 20,000 Hz

1. आवृत्ति माध्यम बदलने पर नहीं बदलती:

• जल में भी आवृत्ति ν = 20,000 Hz रहेगी।

1. जल में नया तरंगदैर्ध्य (λ_जल) ज्ञात करना:

• v_जल = 1480 m/s
• λ_जल = (1480)/(20000) = 0.074 m (7.4 cm)

1. प्रभाव: जल में डुबाने पर तरंगदैर्ध्य 0.017 m से बढ़कर 0.074 m हो जाता है (लगभग 4.35 गुना बढ़ जाता है)।

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 पद-दर-पद हल:
• तरंगदैर्ध्य (λ) = 1 cm = 0.01 m
• वेग (v) = 330 m/s
• सूत्र: ν = (v)/(λ)
• गणना: ν = (330)/(0.01) = 33,000 या 33 kHz
• श्रव्यता: मनुष्य के सुनने की क्षमता 20 Hz से 20,000 Hz तक होती है। चूंकि 33,000 Hz श्रव्य सीमा (20 kHz) से अधिक है, अतः यह एक पराश्रव्य तरंग है और सामान्य मनुष्य इसे नहीं सुन सकता।