Skip to main content
Chapter 11 of 12
NCERT Solutions

Sound

Madhya Pradesh Board · Class 9 · Science

NCERT Solutions for Sound — Madhya Pradesh Board Class 9 Science.

29 questions25 flashcards5 concepts

Interactive on Super Tutor

Studying Sound? Get the full interactive chapter.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan — built for ncert solutions and more.

1,000+ Class 9 students started this chapter today

A diagram illustrating a transverse wave, such as ripples on a water surface, showing particles oscillating perpendicular to the direction of wave propagation.
Super Tutor

Learn better with visuals Super Tutor has hundreds of illustrations like this across every chapter — all free to try.

Get started
17 Questions Solved · 1 Section

9 worked solutions below. Unlock all 17 free in Super Tutor

Exercises

1What is sound and how is it produced?Show solution
Sound:
Sound is a form of energy that produces the sensation of hearing in our ears. It is a mechanical wave that requires a material medium (solid, liquid, or gas) for its propagation.

How sound is produced:
Sound is produced by the vibration of objects. When an object vibrates, it causes the particles of the surrounding medium to vibrate. These vibrations travel through the medium in the form of a wave, which we perceive as sound.

Example: When we pluck a stretched rubber band, it vibrates and produces sound. When a drum is beaten, its membrane vibrates and produces sound. The vibrating object acts as the source of sound.

Not sure why a step works? check your working in Super Tutor

2Describe with the help of a diagram, how compressions and rarefactions are produced in air near a source of sound.Show solution
Production of Compressions and Rarefactions:

Consider a vibrating tuning fork as the source of sound.

- When the prong of the tuning fork moves outward (forward), it pushes the air particles in front of it. These particles get crowded together, forming a region of high pressure called compression (C).
- When the prong moves inward (backward), it creates a region where air particles are spread apart, forming a region of low pressure called rarefaction (R).
- As the tuning fork continues to vibrate, a series of compressions and rarefactions are produced alternately in the air.

Diagram (description):
SourceCRCRC\text{Source} \rightarrow \underbrace{||||}_{C} \quad \underbrace{\quad\quad}_{R} \quad \underbrace{||||}_{C} \quad \underbrace{\quad\quad}_{R} \quad \underbrace{||||}_{C} \rightarrow

Where CC = Compression (region of high pressure, particles close together) and RR = Rarefaction (region of low pressure, particles spread apart).

These compressions and rarefactions propagate through the air as a longitudinal sound wave.

Not sure why a step works? check your working in Super Tutor

3Why is sound wave called a longitudinal wave?Show solution
Sound wave is called a longitudinal wave because the particles of the medium vibrate in the same direction as the direction of propagation of the wave.

Explanation:
When a sound wave travels through air, the air particles are displaced back and forth along the direction in which the wave is moving. This creates alternating regions of compressions (high pressure) and rarefactions (low pressure) along the direction of wave propagation.

Since the direction of particle vibration is parallel (along the same line) to the direction of wave propagation, sound waves are classified as longitudinal waves.

This is in contrast to transverse waves (like light waves), where particle vibration is perpendicular to the direction of wave propagation.

Not sure why a step works? check your working in Super Tutor

4Which characteristic of the sound helps you to identify your friend by his voice while sitting with others in a dark room?Show solution
The characteristic of sound that helps us identify a friend by his voice in a dark room is quality (or timbre) of sound.

Explanation:
Even if two sounds have the same pitch (frequency) and the same loudness (amplitude), they can differ in quality. Quality of sound depends on the waveform of the sound, which is determined by the number and relative intensities of the overtones (harmonics) present in the sound.

Every person has a unique vocal structure, so the sound produced by each person has a unique waveform (quality/timbre). This unique quality allows us to distinguish one person's voice from another even in darkness.

Not sure why a step works? check your working in Super Tutor

5Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?Show solution
Given/Known:
- Flash (light) and thunder (sound) are produced simultaneously during lightning.
- Speed of light =3×108 m s1= 3 \times 10^8 \ \text{m s}^{-1}
- Speed of sound in air 344 m s1\approx 344 \ \text{m s}^{-1}

Reason:
The speed of light is enormously greater than the speed of sound. Light travels at 3×108 m s13 \times 10^8 \ \text{m s}^{-1}, while sound travels at only about 344 m s1344 \ \text{m s}^{-1} in air.

Therefore, light from the flash reaches our eyes almost instantaneously, while sound (thunder) takes a much longer time to travel the same distance. This is why we see the flash first and hear the thunder a few seconds later.

Conclusion: The time gap between seeing the flash and hearing the thunder is due to the vast difference in the speeds of light and sound.

Not sure why a step works? check your working in Super Tutor

6A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in air as 344 m s⁻¹.Show solution
Given:
- Speed of sound in air, v=344 m s1v = 344 \ \text{m s}^{-1}
- Frequency f1=20 Hzf_1 = 20 \ \text{Hz}
- Frequency f2=20 kHz=20,000 Hzf_2 = 20 \ \text{kHz} = 20{,}000 \ \text{Hz}

Formula:
λ=vf\lambda = \frac{v}{f}

**For f1=20 Hzf_1 = 20 \ \text{Hz}:**
λ1=34420=17.2 m\lambda_1 = \frac{344}{20} = 17.2 \ \text{m}

**For f2=20,000 Hzf_2 = 20{,}000 \ \text{Hz}:**
λ2=34420,000=0.0172 m=1.72 cm\lambda_2 = \frac{344}{20{,}000} = 0.0172 \ \text{m} = 1.72 \ \text{cm}

Result:
- Wavelength corresponding to 20 Hz =17.2 m= 17.2 \ \text{m}
- Wavelength corresponding to 20 kHz =0.0172 m= 0.0172 \ \text{m} (or 1.72 cm1.72 \ \text{cm})

Not sure why a step works? check your working in Super Tutor

7Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in air and in aluminium to reach the second child.Show solution
Given:
- Speed of sound in air, vair=346 m s1v_{\text{air}} = 346 \ \text{m s}^{-1} (standard value used in NCERT)
- Speed of sound in aluminium, vAl=6420 m s1v_{\text{Al}} = 6420 \ \text{m s}^{-1} (standard value from NCERT table)
- Let the length of the aluminium rod =d= d

Formula:
Time=DistanceSpeed\text{Time} = \frac{\text{Distance}}{\text{Speed}}

Time taken by sound through air:
tair=dvair=d346t_{\text{air}} = \frac{d}{v_{\text{air}}} = \frac{d}{346}

Time taken by sound through aluminium:
tAl=dvAl=d6420t_{\text{Al}} = \frac{d}{v_{\text{Al}}} = \frac{d}{6420}

Ratio:
tairtAl=d/346d/6420=642034618.5518.5\frac{t_{\text{air}}}{t_{\text{Al}}} = \frac{d/346}{d/6420} = \frac{6420}{346} \approx 18.55 \approx 18.5

tairtAl18.5:1\boxed{\frac{t_{\text{air}}}{t_{\text{Al}}} \approx 18.5 : 1}

This means sound takes about 18.5 times longer to travel through air than through aluminium for the same distance.

Not sure why a step works? check your working in Super Tutor

8The frequency of a source of sound is 100 Hz. How many times does it vibrate in a minute?Show solution
Given:
- Frequency of sound, f=100 Hzf = 100 \ \text{Hz}
- Time =1 minute=60 seconds= 1 \ \text{minute} = 60 \ \text{seconds}

Concept:
Frequency is defined as the number of vibrations (oscillations) per second.
f=Number of vibrationsTime (in seconds)f = \frac{\text{Number of vibrations}}{\text{Time (in seconds)}}

Calculation:
Number of vibrations in 1 minute=f×t=100 Hz×60 s\text{Number of vibrations in 1 minute} = f \times t = 100 \ \text{Hz} \times 60 \ \text{s}
=6000 vibrations= 6000 \ \text{vibrations}

Answer: The source vibrates 6000 times in one minute.

Not sure why a step works? check your working in Super Tutor

9Does sound follow the same laws of reflection as light does? Explain.Show solution
Yes, sound follows the same laws of reflection as light does.

Laws of Reflection of Sound:

1. First Law: The angle of incidence of the sound wave is equal to the angle of reflection.
i=r\angle i = \angle r

2. Second Law: The incident sound wave, the reflected sound wave, and the normal to the reflecting surface at the point of incidence — all lie in the same plane.

Explanation:
Just like light, when sound waves strike a hard surface (like a wall, cliff, or building), they bounce back. The direction of the reflected sound wave obeys the same geometrical rules as reflected light. This is why we hear echoes — the reflected sound follows the same laws as reflected light.

Practical evidence: The phenomenon of echo and the working of a megaphone, stethoscope, and soundboards all demonstrate that sound obeys the laws of reflection.

Not sure why a step works? check your working in Super Tutor

10When a sound is reflected from a distant object, an echo is produced. Let the distance between the reflecting surface and the source of sound production remains the same. Do you hear echo sound on a hotter day?
11Give two practical applications of reflection of sound waves.
12A stone is dropped from the top of a tower 500 m high into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10 m s⁻² and speed of sound = 340 m s⁻¹.
13A sound wave travels at a speed of 339 m s⁻¹. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?
14What is reverberation? How can it be reduced?
15What is loudness of sound? What factors does it depend on?
16How is ultrasound used for cleaning?
17Explain how defects in a metal block can be detected using ultrasound.

8 more solved questions in Sound

Every remaining exercise is solved step by step in Super Tutor, plus practice quizzes and flashcards for this chapter. Free to start.

Stuck on a step?

Ask Super Tutor AI to explain any solution on this page in a simpler way — free, 24x7.

Ask a Doubt Free

Frequently Asked Questions

What are the important topics in Sound for Madhya Pradesh Board Class 9 Science?
Sound covers several key topics that are frequently asked in Madhya Pradesh Board Class 9 board exams. Focus on the core concepts listed on this page and practise related questions to build confidence.
How to score full marks in Sound — Madhya Pradesh Board Class 9 Science?
Understand the core concepts first, then work through the 29 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
Where can I get free NCERT Solutions for Sound Class 9 Science?
This page has free step-by-step NCERT Solutions for every exercise question in Sound (Madhya Pradesh Board Class 9 Science) — written the way examiners award marks: given, formula, working, answer.

Sources & Official References

Content is aligned to the official syllabus. Refer to the board website for the latest curriculum.

For serious students

Get the full Sound chapter — for free.

Quizzes, flashcards, AI doubt-solver and a step-by-step study plan for Madhya Pradesh Board Class 9 Science.