Sound Waves Characteristics and Applications Class 9 Question Answer Science Exploration Chapter 10

Students can use Exploration Class 9 Science Solutions Chapter 10 Sound Waves Characteristics and Applications Question Answer NCERT Solutions as a quick reference guide.

Class 9 Science Exploration Chapter 10 Question Answer

Class 9 Science Ch 10 Sound Waves Characteristics and Applications Question Answer

Sound Waves Characteristics and Applications Class 9 Questions and Answers (Exercise)

Revise, Reflect, Refine (NCERT Textbook Page No. 204 – 206)

Question 1.
Which observation best supports the idea that sound is a mechanical wave?
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
Answer:
(ii) Sound needs a medium to propagate. Mechanical waves are defined as waves that require a material medium for propagation. The fact that sound cannot travel in vacuum (as shown by the bell jar experiment) directly supports this.

Question 2.
For a sound wave propagating in a medium, increasing its frequency will increase its
(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
Answer:
(iii) Number of compressions per second.
Frequency is the number of oscillations (compressions) per second. Increasing frequency directly increases the number of compressions passing a point per second. Speed is a property of the medium (not frequency, in most media). Wavelength decreases with increasing frequency (since v = λν and ν is constant). Time period decreases (T = \(\frac {1}{v}\)).

Question 3.
If 20 compressions pass a point in 4 seconds, the frequency is
(i) 80 Hz
(ii) 5 Hz
(iii) 10 Hz
(iv) 0.2 Hz
Answer:
Here,
number of compressions = 20
and Time = 4 s
Frequency = \(\frac{\text { Number of compressions }}{\text { Time }}\)
Hence, correct option is (ii) 5 Hz.

Question 4.
In a room, the reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify your answer.
Answer:
It will produce reverberation, not an echo.

For an echo, the time gap between the original and reflected sound must be at least 0.1 s for the brain to perceive them as separate sounds. Since 0.05 s < 0.1 s, the brain cannot distinguish the reflected sound from the original. Instead, the reflected sound mixes with the original, prolonging and persisting the original sound-this is reverberation (which occurs when reflections arrive with a time difference less than 0.05 s is for multiple reflections; here 0.05 s is exactly the boundary – but since it is less than 0.1 s, the sounds are not heard separately, making it reverberation).

Question 5.
Graphs representing two sound waves are given in Fig. below. If the scales on the X and Y axes of the two graphs are the same, which of the two sound waves has
(i) greater wavelength, and
(ii) smaller amplitude?

Answer:
(i) Greater Wavelength:
Graph (a) has a greater wavelength.
Wavelength is the horizontal distance between two consecutive crest. In graph (a), the waves are more spread out, meaning there is more distance between crest compared to graph (b).

(ii) Smaller Amplitude:
Graph (a) has a smaller amplitude.
Amplitude is the maximum height of the wave from the center line (density axis). The crest of wave
(a) is lower than the crest of wave (b).

Question 6.
The sound waves emitted by three sources A, B and C are represented in Fig. (below). If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Answer:
Frequency is inversely related to wavelength (v = λν, speed is constant).
Higher frequency means shorter wavelength (more crests per unit distance).
Lower frequency means longer wavelength (fewer crests per unit distance).

From figure (three overlapping curves):
A (maximum frequency) → the curve with the shortest wavelength (most closely spaced crests).
C (minimum frequency) → the curve with the longest wavelength (most widely spaced crests).
B → the curve with wavelength between A and C

Question 7.
Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm.
Answer:
Here,
X-axis: Distance (in cm) and
Y-axis: Density
A horizontal dashed line at average density (e.g., at the middle).

A sinusoidal curve where:
The maximum density (crest) is 3 units above the average density line.
The minimum density (trough) is 3 units below the average density line.
The distance between two consecutive crests (or troughs) = 4 cm the wavelength.

 

Question 8.
In a movie, while showing the explosion of a spacecraft in space, a flash of light is shown along with sound at the same time. What are the errors in this depiction?
Answer:
There are two errors:
Sound cannot travel in space (vacuum):
Outer space has near vacuum – there is no medium for sound to propagate. Sound is a mechanical wave and requires a material medium. So, no sound from the explosion should be heard.

Light and sound travel at vastly different speeds:
Light travels much faster (300,000 km s^-1) than sound (approximately 340 m s^-1 in air). So they would not be perceived simultaneously from any significant distance. The flash of light would be seen far before any sound (if it could exist) would be heard.

Question 9.
A source produces a sound wave of wavelength 3.44 m. If the wave travels with a speed of 344 ms^-1 find its time period.
Answer:
We know,
v = λν
v = \(\frac{v}{\lambda}=\frac{344}{3.44}\) = 10 Hz
Now T = \(\frac{1}{v}=\frac{1}{100}\) = 0.01 s

Question 10.
A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s. If ultrasonic wave travels at 1525 ms^-1 in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?
Answer:
Total time = 5 s;
time to reach the sunken ship = \(\frac {5}{2}\) = 2.5 s
Distance = speed × time
= 1525 × 2.5 = 3812.5 m
≈ 3812.5 m (approximately 3.8 km)

Question 11.
A vehicle is fitted with an ultrasonic distance sensor as part of parking assistance system which provides echolocation, while the driver is reversing the vehicle. It emits ultrasonic wave (about 40 kHz) which is reflected by the obstacle. When the warning beep starts sounding at a distance of 1.2 m from the obstacle, how much time is taken by ultrasonic wave to travel to the obstacle and come back? Assume the speed of ultrasonic wave in air to be 345 m s^-1.
Answer:
Here, distance to obstacle = 1.2 m
Speed = 345 m/s
Wave travels to the obstacle and back,
so total distance = 2 × 1.2 = 2.4 m
Time = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac {2.4}{345}\)
≈ 0.00696 s ≈ 0.007 s.

Question 12.
The speed of sound in air is about 331 ms^-1 at 0 °C and nearly 344 ms^-1 at 22 °C. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m, if the air temperature changes from 22 °C to 0 °C? Assume that all other conditions remain unchanged.
Answer:
Given:
Distance = 1720 m
Speed at 22°C = 344 m/s
Speed at 0°C = 331 m/s
Time at 22°C:
t₁ = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac {1720}{344}\) = 5 s

Time at 0°C:
t₂ = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac {1720}{331}\) s

Extra time taken: ∆t = t₂ – t₁
= \(\frac {1720}{331}\) – 5
= \(\frac {1720-1655}{331}\)
= \(\frac {65}{331}\) s

Question 13.
The variation of density of medium for a sound wave propagating with a speed of 340 m s~’ is shown in Fig. (below). Calculate the wavelength and frequency of the sound wave.

Answer:
In Fig. the distance from the center of one compression to the center of the next compression represents one full wavelength. Looking at the diagram: The distance marked 8 cm spans exactly two full wavelengths (two sets of compressions and rarefactions).
2λ = 8 cm
⇒ λ = \(\frac {8}{2}\)
= 4 cm = 0.04 m
For frequency, using the wave equation
v = ν × λ
v = \(\frac{v}{\lambda}\)
= \(\frac {340}{0.04}\)
= 8500 Hz

Question 14.
The graphical representation of two sound waves A and B propagating at the same speed of 345 m s~’ is shown in Fig. below. What is the wavelength of each of them? Also, calculate their frequencies.

Answer:
From Fig. (x-axis up to 5.0 cm):
Wave A (red curve):
One complete wave spans approximately 2.5 cm.
λ_A = 2.5 cm = 0.025 m
v_A = \(\frac{v}{\lambda}\)
= \(\frac {345}{0.025}\) = 13800 Hz

Wave B (blue curve):
One complete wave spans approximately 5.0 cm.
λ_B = 5.0 cm = 0.05 m
v_B = \(\frac{v}{\lambda}\)
= \(\frac {345}{0.05}\) = 6900 Hz.

Question 15.
Two identical sound sources are placed at A and B – one in air and one submerged in water (Fig. below). Both produce sounds at the same time, which travel horizontally to the vertical side of the cliff and come back. If the time taken by the sound to return to A is 4.5 times than that of B, what is the ratio between the speeds of sound in air and water?

Answer:
Let the distance to the cliff be the same for both sources = d
(Sound travels to the cliff and back, so total distance = 2d)
For source A (in air):
t_air = \(\frac{2 d}{v_{\text {air }}}\)
For source B (in water):
t_water = \(\frac{2 d}{v_{\text {water }}}\)
Given:
t_air = 4.5 × t_water

Class 9 Science Chapter 10 Sound Waves Characteristics and Applications Question Answer (InText)

Think It Over (NCERT Textbook Page No. 184)

Question 1.
Two astronauts are repairing the arm of a space station together during a spacewalk. Can they talk to each other and hear the sounds of metal clanking as they do on the Earth?
Answer:
No, they cannot hear sounds directly because space is a vacuum and sound requires a medium to travel. They communicate using radio devices in their suits, which transmit signals through electromagnetic waves.

Question 2.
How do most bats use sound to locate their prey in the dark at night?
Answer:
Bats use echolocation. They emit ultrasonic sounds which reflect off objects and return as echoes. By detecting these echoes, bats can locate the position, distance, and movement of their prey.

Pause and Ponder (NCERT Textbook Page No. 186)

Question 1.
Explore various ways of producing sound.
Answer:
Sound is produced when objects vibrate. Different ways to produce vibrations include:
(a) By Striking (Hitting) – When struck, the drum vibrates and produces sound.
(b) By Plucking – Plucking strings of guitar makes them vibrate.
(c) By Blowing – Blowing air causes the air column inside the flute to vibrate.
(d) By Rubbing – Rubbing strings of violin with a bow produces vibrations.
(e) By Shaking – Shaking rattles cause parts inside to vibrate.

Question 2.
Make a list of different types of musical instruments and identify their vibrating parts which produce sound.
Answer:
Musical instruments and their vibrating parts:

Musical Instrument	Vibrating Part
Guitar	Stretched strings
Sitar	Stretched strings
Violin	Stretched strings
Tabla	Stretched membrane (skin)
Drum	Stretched membrane
Flute	Air column inside the instrument
Trumpet	Air column
Harmonium	Reeds (metal strips)
Bell	Entire body vibrates
Piano	Strings

Pause and Ponder (NCERT Textbook Page No. 188)

Question 3.
Assertion (A): We cannot hear the sound of a bell ringing in a closed jar after most of the air is pumped out.
Reason (R): Sound requires a medium to travel. Choose the correct statement:
(i) Both A and R are true, but R is not the correct explanation of A
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:
(ii) Both A and R are true, and R is the correct explanation of A.
Explanation:
When air is pumped out of the jar, it becomes nearly a vacuum, so the sound of the bell cannot be heard because sound requires a material medium (like air) to travel.

Pause and Ponder (NCERT Textbook Page No. 191)

Question 4.
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.
Choose the correct statement:
(i) Both A and R are true, but R is not the correct explanation of A.
(ii) Both A and R are true, and R is the correct explanation of A.
(iii) A is true, but R is false.
(iv) A is false, but R is true.
Answer:
(iii) A is true, but R is false.
Explanation:
Compressions and rarefactions move through the medium but particles do not travel forward with the wave. They only vibrate back and forth about their mean position.

Pause and Ponder (NCERT Textbook Page No. 192)

Question 5.
When sound travels from a tuning fork to your ear, which of the following actually reaches your ear?
(i) Air particles near the tuning fork
(ii) Energy carried by sound waves
(iii) The tuning fork material
(iv) A continuous stream of compressed air
Answer:
(ii) Energy carried by sound waves. In sound propagation, it is the energy that is transferred through the medium. The air particles only vibrate about their mean positions and do not travel to the ear.

Pause and Ponder (NCERT Textbook Page No. 193)

Question 6.
The variation of density of the medium for two sound waves is shown in Fig. (a) and (b) below. Label compression and rarefaction by C and R on it. In the graph given in Fig. (c) and (d) below, label the axes and draw the curves corresponding to Fig. (a) and (b).

Answer:

Pause and Ponder (NCERT Textbook Page No. 195)

Question 7.
Conduct Activity 10.1 (See NCERT Page 185) once again with a thick rubber hand and then with a thin rubber hand. Does the thin rubber hand vibrate faster than the thick rubber band? If yes, how do the frequency and time period of the sound produced by the thin rubber band differ from that of the thick rubber hand?
Answer:
Yes, the thin rubber band vibrates faster than the thick rubber band. Since frequency is higher when vibrations are faster.

The frequency of the sound produced by the thin rubber band is higher than that of the thick rubber band. The time period of the thin rubber band’s sound is
shorter (since T = \(\frac{1}{v}\) and frequency is higher, time period is smaller).

Question 8.
If the frequency of a sound wave produced by an oscillating piston of a long tube filled with air is 20 Hz, then how many oscillations does the piston complete per minute?
Answer:
Given: Frequency (v) = 20 Hz (i.e., 20 oscillations per second)
Oscillations in 1 minute = 20 × 60 = 1200 oscillations
The piston completes 1200 oscillations per minute

Question 9.
For the sound wave represented by the graph shown in Fig. (below), what is half of its wavelength?

Answer:
In given fig., the x-axis shows distance up to 4.5 cm with markings at 1.5 and 3.0 cm.
One complete wave (crest to crest or trough to trough) spans 3.0 cm, so the wavelength λ = 3.0 cm.
Half of wavelength = \(\frac {3.0}{2}\) = 1.5 cm.

Pause and Ponder (NCERT Textbook Page No. 197)

Question 10.
Table (below) shows the speed of sound in a few media at atmospheric pressure.

Compare the speeds in different media by finding the ratio of
(i) the speed of sound in water with respect to the speed in the air.
(ii) the speed of sound in steel with respect to the speed in the water.
Answer:
(i) Here,
\(\frac{\text { Speed in water }}{\text { Speed in air }}=\frac {1500}{340}\)
= \(\frac {150}{34}=\frac {75}{17}\)
Hence, sound travels approximately 4.41 times faster in water than in air.

(ii) To find this ratio, we divide the speed of sound in steel by the speed of sound in water:
\(\frac{\text { Speed in steel }}{\text { Speed in water }}=\frac {5000}{1500}\)
= \(\frac {50}{15}=\frac {10}{3}\)
Hence, sound travels approximately \(\frac {10}{3}\) times faster in steel than in water.

Pause and Ponder (NCERT Textbook Page No. 198)

Question 11.
Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. below). Gunjan places her ear over the fence and her friend knocks the fence with a metal object. Using the values of the speed of sound in steel and air given in Table (Above Table), calculate the time difference between the sound that reached Gunjan through the air and the steel. Would it have been possible for her to distinguish between the two sounds? (The time interval between two sounds must be at least 0.1 s to be heard separately.)

Answer:
Given:
Distance = 340 m
Speed of sound in air ≈ 340 m/s
Speed of sound in steel ≈ 5000 m/s
Time taken through air,
t_air = \(\frac{\text { Distance }}{\text { Speed }}\)
= \(\frac {340}{340}\) = 1 s
Time taken through steel,
t_steel = \(\frac {340}{5000}\) = 0.068 s
Time difference,
∆t = t_air – t_steel
= 1 – 0.068 = 0.932 s
Since 0.943 s > 0.1 s.
Yes, Gunjan can clearly hear the two sounds separately.

Pause and Ponder (NCERT Textbook Page No. 201)

Question 12.
An experiment is being set up that requires echoes to arrive at least 0.2 s after the emission of sound. What minimum distance should a reflecting surface be placed at? Assume the speed of sound to be 343 m s^-1.
Answer:
Here, minimum time for echo = 0.2 s
Speed of sound = 343 m/s
Echo time includes to-and-fro travel of sound.
So total distance = 2 × distance to reflecting surface
Total distance = v × t
= 343 × 0.2 = 68.6 m
Minimum distance to reflecting surface = \(\frac {68.6}{2}\) = 34.3 m

Pause and Ponder (NCERT Textbook Page No. 203)

Question 13.
Sound travels much farther in water than light, and thus, is used for various underwater applications. A sonar signal sent to find the depth of ocean takes 4 s to return. What is the depth of the ocean at that location if the speed of sound in seawater is 1500 ms^-1?
Answer:
We have, speed of sound in seawater = 1500 m/s
Total time for signal to go down and come back = 4 s
So, the sound actually travels twice the depth.
Total distance = speed × time
= 1500 × 4 = 6000 m
Depth = \(\frac {6000}{2}\) = 3000 m
The depth of the ocean is 3000 meters (or 3 km).