Class 9 Science(Exploration) Chapter 10 - Sound Waves: Characteristics and Applications Question Answers: Download Solution PDF

NCERT Solutions for Class 9 Science Chapter 10 Sound Waves: Characteristics and Applications Question Answers PDF is prepared to help students understand the chapter in a clear and simple way. In this NCERT solution, you get well-explained answers to all questions, covering important topics like sound waves, their properties such as frequency, amplitude, wavelength, speed of sound, and applications like echo and sonar. This PDF is useful for daily practice as well as quick revision before exams. It is a helpful resource to strengthen your concepts and score better marks in tests.

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NCERT Solution Class 9 Science Chapter 10 - Sound Waves: Characteristics and Applications Revise, Reflect, Refine Question Answer:

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.

Sound is a mechanical wave because it cannot travel in vacuum. It requires a material medium such as air, water, or solids for its propagation.

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 defined as the number of vibrations or compressions per second. When frequency increases, the number of compressions per second also increases, while the speed of sound in the same medium remains unchanged.

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:
(ii) $\mathbf{5 ~ H z}$.
Frequency is defined as the number of compressions per second.

$
\text { Frequency }=\frac{\text { Number of compressions }}{\text { Time }}=\frac{20}{4}=5 \mathrm{~Hz}
$

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 to be heard, the reflected sound must reach the ear after at least 0.1 s. In this case, the sound returns after 0.05 s, which is less than 0.1 s. Therefore, the reflected sound mixes with the original sound and produces reverberation.

5. Graphs representing two sound waves are given in Fig. 10.30. 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

From the given graphs, both waves are plotted on the same scale.

(i) Greater wavelength

The wave in graph (a) has greater wavelength.

This is because it completes fewer cycles over the same distance, so the distance between two consecutive compressions (or crests) is larger.

(ii) Smaller amplitude

The wave in graph (a) has smaller amplitude.

This is because its maximum displacement from the mean position is less compared to graph (b).

6. The sound waves emitted by three sources A, B and C are represented in Fig. 10.31. If the frequency of A is maximum and C is minimum, identify the corresponding curves, and mark A, B and C on them.

Answer

In the given figure, frequency depends on the number of oscillations in a given distance. The wave with more oscillations has higher frequency, and the wave with fewer oscillations has lower frequency.

The green curve shows the maximum number of oscillations, so it represents A with maximum frequency.

The red curve shows a moderate number of oscillations, so it represents B.

The blue curve shows the least number of oscillations, so it represents C with minimum frequency.

7. Draw a graph to represent a sound wave for which the density amplitude is 3 units and wavelength is 4 cm

Answer

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? The depiction in the movie is incorrect due to the following reasons.

Answer

The depiction in the movie is incorrect due to the following reasons.

Sound is a mechanical wave and requires a material medium to travel. In space, there is no medium, so sound cannot propagate. Therefore, no sound of explosion should be heard.

Also, light travels much faster than sound. Even if sound were possible, the flash of light would be seen before the sound is heard. Showing both at the same time is incorrect.

9. A source produces a sound wave of wavelength 3.44 m . If the wave travels with a speed of $344 \mathrm{~m} \mathrm{~s}^{-1}$ find its time period.

Answer
Given:
Wavelength $\lambda=3.44 \mathrm{~m}$
Speed $v=344 \mathrm{~m} \mathrm{~s}^{-1}$
We know,

$
\begin{gathered}
v=\nu \lambda \\
\nu=\frac{v}{\lambda}=\frac{344}{3.44}=100 \mathrm{~Hz}
\end{gathered}
$

Time period $T=\frac{1}{\nu}$

$
T=\frac{1}{100}=0.01 \mathrm{~s}
$

The time period of the wave is $\mathbf{0 . 0 1 ~} \mathbf{~ s}$.

10. A ship searching for a sunken ship sent a sonar signal and detected an echo after 5 s . If ultrasonic wave travels at $1525 \mathrm{~m} \mathrm{~s}^{-1}$ in seawater, approximately how far down in the ocean is the wreckage of the sunken ship located?

Answer
Given:
Speed of sound in seawater $v=1525 \mathrm{~m} \mathrm{~s}^{-1}$
Time taken for echo $t=5 \mathrm{~s}$

The time given is for the to and fro journey of the sound wave.
So, time taken to reach the wreckage

$
t^{\prime}=\frac{5}{2}=2.5 \mathrm{~s}
$

Distance is given by

$
\text { Distance }=v \times t^{\prime}=1525 \times 2.5=3812.5 \mathrm{~m}
$

The wreckage is approximately $\mathbf{3 8 1 2 . 5 ~ m}$ below the surface of the ocean.

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 $345 \mathrm{~m} \mathrm{~s}^{-1}$.

Answer
Given:
Distance from obstacle $d=1.2 \mathrm{~m}$
Speed of ultrasonic wave $v=345 \mathrm{~m} \mathrm{~s}^{-1}$
The wave travels to the obstacle and comes back, so total distance

$
\text { Total distance }=2 \times 1.2=2.4 \mathrm{~m}
$

Time taken

$
\begin{gathered}
t=\frac{\text { distance }}{\text { speed }}=\frac{2.4}{345} \\
t \approx 0.00696 \mathrm{~s} \approx 0.007 \mathrm{~s}
\end{gathered}
$

The time taken by the ultrasonic wave is approximately 0.007 s .

12. The speed of sound in air is about $331 \mathrm{~m} \mathrm{~s}^{-1}$ at $0^{\circ} \mathrm{C}$ and nearly $344 \mathrm{~m} \mathrm{~s}^{-1}$ at $22^{\circ} \mathrm{C}$. Roughly how much extra time will the sound of thunder take to travel a distance of 1720 m , if the air temperature changes from $22^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ ? Assume that all other conditions remain unchanged.

Answer
Given:
Distance $d=1720 \mathrm{~m}$
Speed at $22^{\circ} \mathrm{C}, v_1=344 \mathrm{~m} \mathrm{~s}^{-1}$
Speed at $0^{\circ} \mathrm{C}, v_2=331 \mathrm{~m} \mathrm{~s}^{-1}$
Time taken at $\mathbf{2 2}^{{\circ}} \mathbf{C}$ :

$
t_1=\frac{1720}{344}=5 \mathrm{~s}
$

Time taken at $0^{\circ} \mathrm{C}$ :

$
t_2=\frac{1720}{331} \approx 5.19 \mathrm{~s}
$

Extra time taken:

$t_2-t_1=5.19-5=0.19 \mathrm{~s}$

The sound of thunder will take approximately 0.19 s more time at $0^{\circ} \mathrm{C}$.

13. The variation of density of medium for a sound wave propagating with a speed of $340 \mathrm{~m} \mathrm{~s}^{-1}$ is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.

Answer
From the figure, the $\mathbf{8 ~ c m}$ shown is the distance covering two wavelengths (two compressions).
So, wavelength

$
\lambda=\frac{8}{2}=4 \mathrm{~cm}=0.04 \mathrm{~m}
$

Given speed of sound,

$
v=340 \mathrm{~m} \mathrm{~s}^{-1}
$

Using the relation,

$
\begin{gathered}
v=\nu \lambda \\
\nu=\frac{v}{\lambda}=\frac{340}{0.04}=8500 \mathrm{~Hz}
\end{gathered}
$
The wavelength is $\mathbf{0 . 0 4 ~ m}$ and the frequency is $\mathbf{8 5 0 0 ~ H z}$.

14. The graphical representation of two sound waves A and B propagating at the same speed of $345 \mathrm{~m} \mathrm{~s}^{-1}$ is shown in Fig. 10.33. What is the wavelength of each of them? Also, calculate their frequencies.

Answer
From the graph, we measure wavelength as the distance between two successive crests.
For wave A
Wave A completes one full cycle in $\mathbf{2 . 5 ~ c m}$.

$
\lambda_A=2.5 \mathrm{~cm}=0.025 \mathrm{~m}
$

Frequency

$
\nu_A=\frac{v}{\lambda_A}=\frac{345}{0.025}=13800 \mathrm{~Hz}
$

For wave B
Wave B completes one full cycle in $\mathbf{5 ~ c m}$.

$
\lambda_B=5 \mathrm{~cm}=0.05 \mathrm{~m}
$

Frequency

$
\nu_B=\frac{v}{\lambda_B}=\frac{345}{0.05}=6900 \mathrm{~Hz}
$

15. Two identical sound sources are placed at A and B - one in air and one submerged in water (Fig. 10.34). 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 between source and cliff be $d$.
Since sound goes and comes back, total distance travelled is $2 d$ for both cases.
Time taken $t=\frac{2 d}{v}$
For air (A):

$
t_A=\frac{2 d}{v_{\text {air }}}
$

For water (B):

$
t_B=\frac{2 d}{v_{\mathrm{water}}}
$

Given:

$
\begin{gathered}
t_A=4.5 t_B \\
\frac{2 d}{v_{\text {air }}}=4.5 \times \frac{2 d}{v_{\text {water }}}
\end{gathered}
$

$
\begin{aligned}
\frac{1}{v_{\text {air }}} & =\frac{4.5}{v_{\text {water }}} \\
v_{\text {water }} & =4.5 v_{\text {air }} \\
\frac{v_{\text {air }}}{v_{\text {water }}} & =\frac{1}{4.5}
\end{aligned}
$

Science NCERT Class 9 Chapter 10 - Sound Waves: Characteristics and Applications: Think It Over Question Answer

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, the astronauts cannot talk to each other or hear the sound of metal clanking directly in space.

This is because sound is a mechanical wave and it requires a material medium (like air) to travel. In space, there is no air or medium, so sound cannot propagate.

However, astronauts can communicate using radio signals through their space suits, as radio waves can travel through vacuum.

2. How do most bats use sound to locate their prey in the dark at night?

Answer:

Most bats use a method called echolocation to locate their prey in the dark.

They produce high-frequency sound waves (ultrasound) which travel through the air and get reflected back after hitting objects like insects. The bats listen to these returning echoes and, from the time delay and intensity of the sound, they can determine the position, distance, and size of the prey.

This helps them hunt accurately even in complete darkness.

Chapter 10 NCERT Solution Class 9 Science Chapter 10 - Sound Waves: Characteristics and Applications and Ponder Questions and Answers:

This section provides simple and clear answers to the Pause and Ponder questions of the chapter. It helps students think more deeply about the concepts and strengthen their understanding.

NCERT Science Class 9 Chapter 10 Pause and Ponder(Page 186)

1. Explore various ways of producing sound.

Answer:

Sound is produced by vibrating objects.

It can be produced by plucking, striking, blowing, or rubbing objects, and also by the vibration of vocal cords in humans.

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

Answer

Different musical instruments produce sound due to the vibration of their parts.

In string instruments like guitar and violin, sound is produced by vibrating strings. In instruments like tabla and drum, sound is produced by a vibrating stretched membrane. In wind instruments such as flute, sound is produced by the vibration of the air column. In harmonium, vibrating reeds produce sound, while in a bell, the metal body vibrates to produce sound.

Thus, different instruments produce sound through different vibrating parts.

NCERT Solution Science Class 9 Chapter 10 Pause and Ponder(Page 188)

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.

The assertion is true because when air is removed from the jar, there is no medium left for sound to travel.

The reason is also true and correctly explains the assertion, since sound requires a material medium for propagation.

Class 9 Chapter 10 NCERT Solution Science Pause and Ponder(Page 191)

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.

The assertion is true because compressions and rarefactions travel through the medium as the sound wave propagates.

The reason is false because the particles of the medium do not move forward continuously with the wave. They only oscillate about their mean positions.

Class 9 Chapter 10 Science NCERT Solution Pause and Ponder(Page 192)

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.

When sound travels, the particles of the medium only vibrate about their mean positions. It is the energy of the sound wave that travels from the source (tuning fork) to the ear, not the particles themselves.

Science Class 9 Chapter 10 NCERT Solution Pause and Ponder(Page 193)

6. The variation of density of the medium for two sound waves is shown in Fig. 10.17 (a) and (b). Label compression and rarefaction by C and R on it. In the graph given in Fig. 10.17 (c) and (d), label the axes and draw the curves corresponding to Fig. 10.17 (a) and (b).

Answer:

NCERT Solution Science Class 9 Chapter 10 Pause and Ponder(Page 195)

7. Conduct Activity 10.1 once again with a thick rubber band and then with a thin rubber band. Does the thin rubber band 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 band?

Answer

Yes, the thin rubber band vibrates faster than the thick rubber band.

The frequency of the sound produced by the thin rubber band is higher, while its time period is smaller compared to the thick rubber band.

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 $\nu=20 \mathrm{~Hz}$
Frequency means number of oscillations per second.
So, oscillations in 1 second $=20$
Oscillations in 1 minute (60 seconds):

$
20 \times 60=1200
$

The piston completes $\mathbf{1 2 0 0}$ oscillations per minute.

9. For the sound wave represented by the graph shown in Fig. 10.19, what is half of its wavelength?

Answer
From Fig. 10.19, the $x$-axis shows distance up to 4.5 cm with markings at 1.5 cm and 3.0 cm . One complete wave (from crest to crest or trough to trough) spans 3.0 cm .

Therefore, the wavelength

$
\lambda=3.0 \mathrm{~cm}
$

Half of the wavelength is

$
\frac{\lambda}{2}=\frac{3.0}{2}=1.5 \mathrm{~cm}
$

Science NCERT Solution Class 9 Chapter 10 Pause and Ponder(Page 197)

10. Table 10.1 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
Given:
Speed of sound in air $=340 \mathrm{~m} \mathrm{~s}^{-1}$
Speed of sound in water $=1500 \mathrm{~m} \mathrm{~s}^{-1}$
Speed of sound in steel $=5000 \mathrm{~m} \mathrm{~s}^{-1}$
(i) Ratio of speed in water to speed in air:

$
\frac{1500}{340} \approx 4.4
$

So, the ratio is approximately $4.4: 1$
(ii) Ratio of speed in steel to speed in water:

$
\frac{5000}{1500}=\frac{10}{3} \approx 3.33
$

So, the ratio is approximately $3.33: 1$

11. Two friends are standing along a steel fence at a distance of 340 m from each other (Fig. 10.23). 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 10.1, 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 $=340 \mathrm{~m}$
Time taken through steel:

$
\frac{340}{5000}=0.068 \mathrm{~s}
$

Time taken through air:

$
\frac{340}{340}=1.0 \mathrm{~s}
$

Time difference:

$
1.0-0.068=0.932 \mathrm{~s}
$

Since $0.932 \mathrm{~s}>0.1 \mathrm{~s}$, Gunjan would be able to distinguish between the two sounds.

Science NCERT Solution Class 9 Chapter 10 Pause and Ponder(Page 201)

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 $343 \mathrm{~m} \mathrm{~s}^{-1}$.

Answer
Given:
Speed of sound $v=343 \mathrm{~m} \mathrm{~s}^{-1}$
Time for echo $t=0.2 \mathrm{~s}$
For echo, sound travels to the surface and back:

$
\text { Total distance }=2 d
$

Using $v=\frac{\text { distance }}{\text { time }}:$

$
\begin{gathered}
2 d=v \times t=343 \times 0.2=68.6 \\
d=\frac{68.6}{2}=34.3 \mathrm{~m}
\end{gathered}
$
The reflecting surface should be placed at a minimum distance of $\mathbf{3 4 . 3 ~ m}$.

NCERT Solution Class 9 Science Chapter 10 Pause and Ponder(Page 203)

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 $1500 \mathrm{~m} \mathrm{~s}^{-1}$ ?

Answer
Given:
Speed of sound $v=1500 \mathrm{~m} \mathrm{~s}^{-1}$
Time $t=4 \mathrm{~s}$
Sound travels to the ocean floor and back, so:
Total distance $=2 d$

$
\begin{gathered}
2 d=v \times t=1500 \times 4=6000 \\
d=\frac{6000}{2}=3000 \mathrm{~m}
\end{gathered}
$

The depth of the ocean is 3000 m .

NCERT Class 9 Science Chapter 10 - Sound Waves: Characteristics and Applications Activity Question Answer:

Class 9 Chapter NCERT Activity 10.1 : Let us explore - Vibrating Rubber Band

1. Take a cardboard box with one side open and a rubber band.
2. Stretch the rubber band across the open side of the box (Fig. 10.2).

3. Holding the box steady with one hand, pluck the rubber band with a finger. Do you hear any sound?
4. Pluck the rubber band again and watch it carefully. Is it vibrating?
5. Wait till the rubber band stops vibrating. Do you still hear the sound?
6. Change the tension in the rubber band by stretching it more or loosening it slightly and plucking it each time. Does the sound change? What changes do you notice?
7. Remove the rubber band from the box. Stretch it between two fingers and pluck it near your ear. Is the sound still produced? Is it as loud as before?

Observation

As long as the rubber band is stretched and vibrating, sound is produced. When the vibration stops, the sound also stops. When the tension in the rubber band is increased, the sound produced has a higher pitch. When the rubber band is removed from the box and plucked in open air, the sound is produced but it is much softer, because the box acts as a resonator.

Explanation

Sound is produced due to vibrations. Vibration refers to the periodic to and fro motion of an object. The box amplifies the sound due to resonance, which makes the sound louder when the rubber band is placed on it. Changing the tension of the rubber band changes its frequency of vibration, which in turn changes the pitch of the sound.

Conclusion

Sound is produced by vibrating objects. When vibration stops, sound also stops. The object that produces sound is called the source of sound.

Class 9 Science Chapter 10 Activity 10.2 : Let us explore - Tuning Fork

1. Take a tuning fork and a soft rubber pad.
2. Hold the tuning fork by its stem.
3. Strike one of the prongs of the tuning fork gently against the rubber pad (Fig. 10.4b) and bring it close to your ear. Do you hear a sound? (Take care not to strike the tuning fork against a hard surface.)

4. Now, gently touch a water surface with one of the vibrating prongs of the tuning fork. Do you see waves forming on the surface of water?
5. Repeat step 3 a few times while bringing the prongs of the tuning fork near your ear in different orientations. Do you hear the sound?

Observation

When the tuning fork is struck against a rubber pad and held near the ear, a clear sound is heard. When the vibrating prong touches the surface of water, ripples are observed. Sound is heard in all orientations of the tuning fork near the ear.

Explanation

When the tuning fork is struck, its prongs start vibrating. These vibrations disturb the water surface and produce ripples, showing that the prongs are vibrating. Sound is heard in all directions because it spreads outward from the vibrating source in all directions. This confirms that sound is produced by vibrating objects.

Conclusion

A tuning fork is a U-shaped metal bar with prongs that vibrate when struck. It produces nearly a single-frequency sound.

NCERT Science Class 9 Chapter 10 Activity 10.3 : Let us investigate - Sound Through Solids

1. You and your friend stand on opposite sides of a desk in the classroom. Let your friend gently knock or scratch on the desk. Listen carefully to the sound produced with your ear in the air.
2. Now, place your ear against the desk, close your other ear and listen again (as shown in Fig. 10.5). Are you able to hear the sound through the table?

Observation

When the ear is in air, the sound of knocking is heard but it is faint. When the ear is placed directly against the desk, the sound becomes louder and clearer.

Explanation

Sound can travel through solids. It travels faster and more efficiently in solids than in air. The desk acts as a medium that transmits sound vibrations directly from the source to the ear, making the sound louder.

Conclusion

Sound can propagate through solid materials. This shows that solids are good conductors of sound.

NCERT Class 9 Science Chapter 10 Activity 10.4 : Let us investigate

1. Take a large tub or bucket of water filled to the brim and two metal spoons.
2. Tap the spoons against one another and listen to the sound produced (Fig. 10.6a).
3. Now, submerge the two metal spoons in water without touching the sides or bottom of the bucket and tap them against one another again (Fig. 10.6b). Do you again hear the sound produced?

Observation

When spoons are tapped in air, a clear metallic sound is heard. When the spoons are tapped while submerged in water, the sound is still heard, though it travels through water and then air to reach the ears.

Explanation

The sound produced by the spoons travels through water and then through air before reaching the ears. This shows that sound can propagate through liquids. The substance through which sound travels is called a medium. Sound can travel through solids, liquids, and gases.

Conclusion

Sound can travel through liquids such as water. It requires a material medium (solid, liquid, or gas) for propagation and cannot travel through vacuum.

NCERT Solution Class 9 Science Chapter 10 Activity 10.5 : Let us observe

1. Take a slinky and a marker.
2. Make a mark on a turn of the slinky with the marker. Lay out the slinky horizontally on a table or floor.
3. Ask a friend to hold one end of the slinky fixed while you hold the other end keeping the slinky slightly stretched.
4. Give the slinky at your end a sharp push towards your friend and then quickly pull it back again (Fig.
10.8). Do you observe a disturbance created in the slinky which moves towards your friend?
5. Now, push and pull the slinky end multiple times in quick succession. Are a series of disturbances produced in the slinky? Do these disturbances move across the length of the slinky? Does the mark on the slinky move back and forth parallel to the direction of the disturbance?

Observation

When one end of the slinky is pushed and pulled, a disturbance travels along its length. Regions where the coils are closer together and regions where they are spread apart can be observed. A marked coil does not move along the slinky but only oscillates back and forth about its mean position.

Explanation

The regions where coils are closer represent compressions, and the regions where they are spread apart represent rarefactions. The disturbance travels through the slinky, but individual coils only vibrate about their mean positions. This shows the behavior of a longitudinal wave, in which particles vibrate parallel to the direction of wave propagation. Sound waves are longitudinal mechanical waves.

Conclusion

In a sound wave, energy travels through the medium, not the particles. The particles only vibrate about their mean positions.

NCERT Class 9 Science Chapter 10 Activity 10.6 : Let us experiment

1. Take a wide mouthed container, a cellophane or rubber sheet (such as that of a balloon) of size larger than its opening, a loud sound source (such as a metal plate and a beater) and some grains or particles (such as rice, semolina, salt, or chalk powder).
2. Stretch the sheet over the edges of the container tightly and fix it with tape or a rubber band (Fig. 10.14).

3. Sprinkle the grains evenly over the sheet, ensuring they are not clumped together.
4. Produce a loud sound near the bowl without touching it. Observe the grains on the sheet. Does the sound have any effect on the grains?
5. Repeat step 4 with different sources of sounds and observe the effect on the grains. You can try increasing or reducing the volume of sound. Try with different grains.

Do you observe the grains over the sheet move or jump? Why does this happen, even though the source of sound is not touching the sheet or the container?

Observation

The grains placed on the stretched sheet start moving and jumping when a loud sound is produced nearby, even though the sound source is not touching the sheet. When louder sounds are produced, the grains jump higher, while for softer sounds, the grains move very little.

Explanation

Sound is a form of energy. When a sound source vibrates, it transfers energy to the surrounding air. As sound waves travel through the air, they reach the stretched sheet and make it vibrate. These vibrations cause the grains to move. Louder sounds carry more energy, resulting in greater vibrations and higher movement of the grains.

Conclusion

Sound is a form of energy and it transfers energy from the source to the medium as it propagates.

Class 9 Science Chapter 10 Activity 10.7 : Let us experiment (Demonstration Activity)

This activity is recommended to be performed as a classroom group activity facilitated by the teacher.
1. Use a mobile app, such as Phyphox, that can identify frequencies of sounds. Use the 'Audio Spectrum' option that displays the frequency graphically or in hertz (Hz).
2. Try to sing the musical notes ' $\mathrm{Sa}, \mathrm{Re}, \mathrm{Ga}, \mathrm{Ma}, \mathrm{Pa}, \mathrm{Dha}, \mathrm{Ni}, \mathrm{Sa}$ ' one after another, or use a music or tone generating app on another phone to produce those notes. Observe how the frequency changes as each note is produced.
3. Record the approximate frequency values for each musical note.
4. Compare the musical notes by taking the ratio of each frequency with respect to the 'Sa'. Do you observe any pattern?
5. If both voice and mobile-generated notes are used, compare their frequencies for the same musical notes.

Observation

The frequency is lowest for ‘Sa’ and increases gradually for the other notes. Each musical note has a distinct frequency. The higher ‘Sa’ has approximately double the frequency of the lower ‘Sa’, which is called an octave.

Observed Frequency Table

Explanation

Each musical note corresponds to a specific frequency of sound. Higher pitch means higher frequency. The pattern of ratios shows that musical notes follow a definite mathematical relationship. The higher ‘Sa’ has nearly double the frequency of the lower ‘Sa’, forming one octave. Sounds produced by voice or devices for the same note have approximately the same frequency, though slight variations may occur.

Conclusion

Each musical note has a distinct frequency. Higher notes have higher frequencies. The higher octave ‘Sa’ has approximately double the frequency of the lower ‘Sa’.

Chapter 10 NCERT Class 9 Activity 10.8 : Let us experiment (Demonstration Activity)

This activity is recommended to be performed as a classroom group activity facilitated by the teacher.
1. Open a mobile app that can generate sounds.
2. Set the frequency to $\mathbf{1 0 0 ~ H z}$, tap 'play', and listen carefully.
3. Increase the frequency in steps of $\mathbf{1 0 0 ~ H z}$ up to $\mathbf{1 0 0 0 ~ H z}$ and describe how the sound changes.
4. Next, set the frequency to $\mathbf{5 0 ~ H z}$. Reduce the frequency till about $\mathbf{2 0 ~ H z}$ or the point where you cannot hear the sound anymore.

Observation

As the frequency is increased from 100 Hz to 1000 Hz, the sound becomes progressively higher in pitch. At 100 Hz, the sound is deep, while at 1000 Hz, it becomes sharper and shriller. When the frequency is reduced below about 20 Hz, the sound is no longer heard.

Explanation

Pitch is the perception of frequency by the human ear. Higher frequency produces higher pitch, while lower frequency produces lower pitch. The human audible range is from 20 Hz to 20,000 Hz. Sound waves below 20 Hz are called infrasonic waves, and those above 20,000 Hz are called ultrasonic waves. Humans cannot hear these, but some animals can detect them.

Conclusion

Higher frequency corresponds to higher pitch. The human audible range is 20 Hz to 20 kHz. Sounds below 20 Hz are infrasonic and above 20 kHz are ultrasonic, both of which are inaudible to humans.

NCERT Solution For Class 9 Science Chapter 10: Topics

10.1 Production of Sound
10.2 Propagation of Sound
10.3 Sound Waves
10.4 Energy of Sound Waves
10.5 Graphical Representation of a Sound Wave
10.6 Characteristics of a Sound Wave
10.7 Reflection of Sound
10.8 Ultrasonic and Infrasonic Waves, and their Applications