Sound Fundamentals for Numerical Problems
Sound waves are longitudinal waves in which particles of the medium vibrate parallel to the direction of wave propagation. In air, sound travels through alternating compressions (high pressure) and rarefactions (low pressure). For a complete theoretical explanation of these concepts, students may refer to the Sound Class 9 Notes.
This page is part of the sound class 9, science chapter hub which includes notes, numericals, assignments, tests, and revision resources.
Key Wave Parameters
Wavelength ($\lambda$): Distance between two consecutive compressions or rarefactions.
Frequency ($f$): Number of oscillations per second (Hz). The audible range for humans is 20 Hz to 20,000 Hz.
Time Period ($T = \frac{1}{f}$): Time taken for one complete oscillation.
- 20 Hz → $T = 0.05$ s
- 20,000 Hz → $T = 5 \times 10^{-5}$ s
Speed of Sound ($v = f\lambda$): In air at 20°C, $v \approx 340$ m/s. Temperature dependence is given by: $$v \approx 331 + 0.6T$$
Amplitude: Maximum displacement of particles from mean position; it determines loudness.
Wave Equation: $v = f\lambda$
Example: A tuning fork of frequency 512 Hz produces sound in air: $$\lambda = \frac{v}{f} = \frac{340}{512} = 0.664 \text{ m}$$
Sound travels faster in solids than in liquids and gases due to higher elasticity. To understand why sound is always associated with vibrations, read: Why is sound produced by vibrations?
Echo and SONAR Applications
An echo is the reflection of sound and can be heard distinctly only when the reflecting surface is at least 17 m away.
Echo distance formula:
$$d = \frac{vt}{2}$$SONAR (Sound Navigation and Ranging) uses this principle for underwater detection. Students can practise more theory-based and numerical questions from: Sound Class 9 Questions and Answers.
Radar works on a similar principle but uses electromagnetic waves instead of sound.
Wavelength Change Across Media
When sound enters a new medium, its frequency remains constant: $$\frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}$$
Example: Sound of wavelength 0.3 m in air enters helium: $$\lambda_{\text{He}} = \frac{970 \times 0.3}{340} = 0.85 \text{ m}$$
Solved Sound Numericals with Answers
CBSE Numerical Problem Types Covered
- Basic formula-based numericals
- Speed, wavelength and frequency calculations
- Echo and distance–time problems
- Mixed sound and motion problems
For additional exam-oriented practice, students may attempt: Extra Questions of Sound – Class 9.
A. Basic Formula Application based Questions
Question 1. Define one hertz?
Answer
One hertz is defined as one oscillation or cycle per second. It is the unit of frequency.
Question 2. Audible frequency range of a human ear is 20 hertz to 20000 hertz. Express it in terms of time period?
Answer
Time period for 20 Hz = $\frac{1}{f} = \frac{1}{20} = 0.05$ s
Time period for 20,000 Hz = $\frac{1}{f} = \frac{1}{20000} = 5 \times 10^{-5}$ s
Hence, the time period range is 0.05 s to $5 \times 10^{-5}$ s.
Question 3. Derive the relationship between wavelength, frequency and speed of sound?
Answer
Speed = distance/time
or
$v = \frac{\lambda}{T}$
or
$v = \lambda \nu$
Question 4. A person is listening to a tone of 500 Hz sitting at a distance of 450m from the source of the sound. What is the time interval between successive compressions from the source? [Speed of sound in air = 330m/s]
Answer
The time interval between successive compressions from the source is equal to the time period and time period is reciprocal of the frequency. Therefore, it can be calculated as follows:
$T = \frac{1}{F}$
$T = \frac{1}{500}$
$T = 0.002$ s
B. Speed, Wavelength & Frequency based Numerical
Question 5. A tuning fork having frequency 312Hz emits a wave which has a wavelength of 1.10m. Calculate the speed of sound?
Answer
$v = \lambda \nu$
$v = 1.10 \times 312 = 343.2$ m/s
Question 6. A wave is moving in air with a velocity of 340m/s. Calculate the wavelength if its frequency is 512Hz?
Answer
$v = \lambda \nu$
$\lambda = \frac{v}{\nu}$
$\lambda = \frac{340}{512} = 0.664$ m
Question 7. A sound wave has a frequency 2 kHz and wavelength 40 cm. How long will it take to travel 1.6 km?
Answer
$\nu = 2000$ Hz
$\lambda = 0.4$ m
Speed of the sound
$v = \lambda \nu = 2000 \times 0.4 = 800$ m/s
Time taken to cover 1.6 km = $\frac{1600}{800} = 2$ sec
Question 8. Sound waves of wavelength λ travel from a medium in which its velocity is v m/s into another medium in which its velocity is 3v m/s. What is the wavelength of the sound in the second medium. Frequency remains constant?
Answer
$v = \lambda \nu$
$\nu = \frac{v}{\lambda}$
As frequency is constant
$\frac{v_1}{\lambda_1} = \frac{v_2}{\lambda_2}$
$\lambda_2 = \frac{v_2 \times \lambda_1}{v_1} = \frac{3v \times \lambda}{v} = 3\lambda$ m
Question 9. In tank, 10 ripples are produced in one sec. If the distance between a crest and trough is 10cm, find?
(a) Wavelength
(b) Frequency
(c) Velocity of the wave?
Answer
The wavelength of a wave is the distance between two consecutive crests or troughs. Here, the distance between a crest and next trough is half of the wavelength.
λ/2 = 10 cm
λ = 20 cm = 0.2 m
Frequency is the number of oscillations per unit time.
f = 10 Hz
Now, velocity of the wave,
v = λν = 0.2 × 10 = 2 m/s
Question 10. When a wave travels from one medium to another, the wavelength changes but not the frequency. The wavelength of sound disturbance 30 cm in air and of the wave velocity is 340 m/s. What will be the wavelength of this disturbance in Helium? The speed of sound in helium is 970 m/s and 1450 m/s in water?
Answer
v = λν
ν = v/λ
As frequency is constant
vair/λair = vhel/λhel
λhel = (vhel × λair)/vair = (970 × 0.3)/340 = 0.85 m
C. Distance–Time Numerical (Echo, Radar & Relative Speed Problems)
Question 11. A radar signal is received 2 × 10-5 sec. After it was sent and reflected by an aeroplane. How far is the aeroplane if speed of waves is 3 × 108 m/s?
Answer
Given,
Speed of wave, v = 3 × 108 m/s
Time taken for receiving the radar signal
t = 2 × 10-5 sec
Distance travelled by the wave
= v × t = 3 × 108 × 2 × 10-5 = 6 × 103 m
In this time wave has to travel twice the distance between the aeroplane and receiver
Hence, the distance from aeroplane
= 3 × 103 m
Question 12. A boy standing in front of a wall at a distance of 85 m produces 2 claps per second. He notices that the sound of his clapping coincides with the echo. The echo is heard only once when clapping is stopped. Calculate the speed of sound.
Answer
Distance of wall from the boy = 85m
To hear the echo, sound has to travel a distance = 2 × 85 = 170 m
Since 2 claps are produced in one second, each clap is produced after = 1/2 = 0.5 sec
Since echo matches with the second clapping.
Speed of sound = 170/0.5 = 340 m/s
Question 13. An engine is approaching a hill at constant speed. When it is at a distance of 0.9km, it blows a whistle, whose echo is heard by the driver after 5s. If the speed of sound is 340m/s, calculate the speed of the engine.
Answer
Let v be the speed of the engine
Initial distance of engine from cliff = 900m
Distance of engine from cliff when echo is heard = 900 - 5v
Total distance covered by sound = 900 + 900 - 5v = 1800 - 5v
Now
Speed = distance/time
340 = (1800 - 5v)/5
1700 = 1800 - 5v
5v = 100
v = 30 m/s
Question 14. An echo is returned in 6 seconds. What is the distance of reflecting surface from source? [Given that speed of sound is 342 m/s.]
Answer
Given
Time in which echo returned, t = 6 s
Speed of sound, v = 342 m/s
Distance = Speed × Time = 342 × 6 = 2052m
As this distance is twice the distance of reflecting surface from source.
So,
The distance of reflecting surface from source = 2052 / 2 = 1026 m
D. Mixed Concept Challenges (Sound + Motion)
Question 15. A stone is dropped from a 500 m tall building into a pond. When is sound of splash heard at the top? (g=10m/s2, speed of sound in air= 340m/s)?
Answer
Here we need to first find the time for the stone to drop in pond
s = ut + (1/2)gt2
Here s = 500m, u = 0, g = 10m/s2
500 = (1/2) × 10 × t2
500 = 5t2
t2 = 100
t = 10 sec
Now time for the sound to travel to peak of tower
= 500/340 = 1.47 sec
So total time to heard the sound = 10 + 1.47 = 11.47 sec
Question 16. Find the distance of cloud from you when you hear a thunder 3 sec after the lightning is seen. (given the speed of light = 3 × 108 m/s, speed of sound = 330m/s)? Why is lightning seen a few seconds before the thunder is heard during a thunderstorm?
Answer
Speed of light is greater than the speed of sound. Therefore, lightning will reach in no time but the thunder will be heard 3 seconds after the lightning by the man.
We know that the lightning and the sound travels the same distance, let us consider that as 'd'.
Now for light
d = 3 × 108 × t
For sound
d = 330 × (t+3) as sound is heard 3 sec later
Equating both
3 × 108 × t = 330(t+3)
3 × 108 × t = 330t + 990
Since 3 × 108 × t >> 330t, we can ignore 330t
3 × 108 × t ≈ 990
t = 990/(3 × 108)
t = 3.30 × 10-6 s
Therefore
d = 3 × 108 × 3.30 × 10-6 = 990 m
Table Type Questions
Question 17.
Answer
| Sound velocity (m/s) | Frequency (Hz) | Wavelength | Time period (s) | Audible / Ultrasonic / Infrasonic |
|---|---|---|---|---|
| 330 | 550 | 0.6 m | 0.00182 | Audible |
| 34000 | 500 | 68 m | 0.002 | Audible |
| 343.4 | 20200 | 1.7 cm | 0.0000495 | Ultrasonic |
Additional Practice Numerical Problems
Question 1. A person hears an echo from the top of a tower 2.2 s after the sound is produced. How far away is the tower? (Speed of sound = 332 m/s)
Question 2. Determine the minimum distance between the listener and reflector for an echo to be heard distinctly if the speed of sound is (v) m/s.
Question 3. A ship is stationary at a distance of 2800 m from the sea-bed. It sends an ultrasound signal and its echo is heard after 4 s. Find the speed of sound in water.
Question 4. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of ultrasound if:
- Speed of ultrasound = 1.5 km/s
- Frequency = 4.0 MHz
Hints and Answers (1 to 4)
Question 1
Hint: In an echo, sound travels to the tower and back. Total distance = 2d.
Answer: Distance of the tower = 365.2 m
Question 2
Hint: For a distinct echo, the minimum time gap should be 0.1 s.
Answer: Minimum distance = (v × 0.1) / 2 = 0.05v m
Question 3
Hint: Ultrasound travels to the sea-bed and back. Total distance = 2 × 2800 m.
Answer: Speed of sound in water = 1400 m/s
Question 4
Hint: Use the relation v = fλ.
Answer: Wavelength of ultrasound = 3.75 × 10−4 m
Question 5. A ship sends ultrasound that returns from the sea-bed and is detected after 3.22 s. If the speed of ultrasound in seawater is 1530 m/s, calculate the distance of the sea-bed from the ship.
Question 6. A body is vibrating 6000 times in one minute. If the velocity of sound in air is 360 m/s, find:
- (a) Frequency of vibration
- (b) Wavelength of sound produced
Question 7. A sound wave produced by a source travels through air and covers a distance of 72 m in 0.2 s. During this time, the source completes 120 vibrations.
Calculate:
- (a) The velocity of sound in air
- (b) The frequency of the sound produced
- (c) The wavelength of the sound wave
Hints and Answers (5 to 7)
Question 5
Hint: Echo time corresponds to sound travelling to sea-bed and back.
Answer: Distance of sea-bed = 2463.3 m (approximately)
Question 6
Hint: Convert vibrations per minute into frequency per second.
Answer:
(a) Frequency = 100 Hz
(b) Wavelength = 3.6 m
Question 7
Hint: First find the velocity using distance and time. Then calculate frequency from number of vibrations per second. Finally, use the relation v = fλ.
Answer:
(a) Velocity of sound = 360 m/s
(b) Frequency = 600 Hz
(c) Wavelength = 0.6 m
Question 8. Meera stands between two hills and shouts, hearing first echo after 0.5 s and second after 1.0 s. Speed of sound is 340 m/s. Find distance between hills.
Question 9. A stone drops from a 500 m building into a pond. Speed of sound in air is 340 m/s, in water 1500 m/s. Calculate time interval between seeing splash and hearing it (neglect stone fall time for sound, g=10g=10 m/s²).
Question 10. A person fires a gun in the direction of a hill and hears the echo after 5 seconds. He then walks 340 meters closer to the hill and fires again, hearing the echo after 3 seconds. Determine the speed of sound in air.
Hints and Answers (8 to 10)
Question 8
Hint: First echo is from the nearer hill, second from the farther hill.
Answer: Distance between the two hills = 255 m
Question 9
Hint: Time interval = time taken by stone to fall + time taken by sound to travel.
Answer: Time interval = 9.7 s (approximately)
Question 10
Hint: Form two echo-time equations before and after walking closer.
Answer: Speed of sound in air = 340 m/s
Online Practice and Revision
After solving these numericals, students are advised to test their understanding using MCQ-based assessments:
For visual revision and quick recall, refer to: Sound Mind Maps for Class 9.
Key Entities Summary:
- Longitudinal Wave: Parallel particle vibration creating compressions/rarefactions
- Frequency: Oscillations/second (Hz); determines pitch
- Wavelength: Distance between compressions; medium-dependent
- Echo: Reflected sound with $d = vt/2$ calculation
- SONAR: Underwater ranging using sound pulse-echo
- Speed of Sound: 340 m/s in air; varies with temperature and medium
This physics numericals for class 9 sound with answers is prepared keeping in mind the latest syllabus of CBSE. This has been designed in a way to improve the academic performance of the students. If you find mistakes, please do provide the feedback on the mail or comment section below.
This page on physics numericals for Class 9 Sound with answers is prepared strictly as per the latest CBSE syllabus. For other chapters, visit the main Class 9 Science page.