Chapter 10 Sound Waves: Characteristics and Applications class 9th (Exploration) ncert solution

Sound Waves: Characteristics and Applications — Solutions | EduGrown
EduGrown • Class 9 Science • Chapter 10 Solutions
Sound Waves In-text + Exercises Step-wise
Part 1

Think It Over, Activities & Pause-and-Ponder

Think It Over

Opening questions about astronauts in space and bats.

Answer

(a) Astronauts on a spacewalk: No, they cannot hear each other speak or hear metal clanking the way they do on Earth. Outer space is a near-vacuum (no medium), and sound needs a material medium to travel. So they communicate using radio devices fitted in their spacesuits.

(b) How bats locate prey: Bats emit short bursts of ultrasonic waves. These waves reflect off nearby objects and prey, and by sensing the returning echoes the bat works out the position of its prey. This is called echolocation.

Activities 10.1 – 10.8

Key conclusions from the chapter’s activities.

Conclusions
  • 10.1 (rubber band): Sound is produced only while the band vibrates; it stops when the vibration stops. Tightening the band changes the sound (higher pitch).
  • 10.2 (tuning fork): A struck tuning fork produces sound and creates ripples when its prong touches water — showing its prongs vibrate.
  • 10.3 (desk): The knock is heard when the ear is on the desk — sound travels through solids.
  • 10.4 (spoons in water): The tapping is heard through the water too — sound travels through liquids.
  • 10.5 (slinky): A disturbance (compressions and rarefactions) travels along the slinky while each turn only oscillates about its place — a longitudinal wave.
  • 10.6 (grains): Grains jump when a loud sound is made nearby — sound carries energy; louder sound makes them jump higher.
  • 10.7 / 10.8 (frequency apps): ‘Sa’ has the lowest frequency and higher notes have higher frequency; as frequency rises, the pitch rises (sound becomes shriller).
Pause and Ponder • Q1

1. Explore various ways of producing sound.

Answer

Sound is always produced by some vibration. Common ways include: plucking a stretched string (guitar, sitar), striking a metal object (bell, taal), blowing air through a pipe (flute, whistle), beating a stretched membrane (drum, tabla), and the vibration of vocal cords while speaking or singing. Animals like crickets rub their wings/legs to make sound.

Pause and Ponder • Q2

2. List different musical instruments and identify their vibrating parts.

Answer
InstrumentVibrating part
Sitar / Guitar / ViolinStretched strings
Tabla / Drum / MridangamStretched membrane (skin)
Flute (Bansuri) / WhistleAir column inside the pipe
Bell / Taal / ManjiraThe metal body itself
HarmoniumMetal reeds
Pause and Ponder • Q3 • Assertion–Reason

3. A: We can’t hear a bell ringing in a jar after most air is pumped out. R: Sound requires a medium to travel.

(i) Both A and R true, R not the correct explanation
(ii) Both A and R true, and R is the correct explanation of A
(iii) A true, R false
(iv) A false, R true
Answer — (ii)

Both are true. As air is removed, there is almost no medium left in the jar, and since sound needs a medium to propagate, it cannot reach us — so R correctly explains A.

Pause and Ponder • Q4 • Assertion–Reason

4. A: Compressions and rarefactions move through the medium. R: Individual particles of the medium continuously move forward with the wave.

(i) Both true, R not correct explanation
(ii) Both true, R correct explanation
(iii) A is true, but R is false
(iv) A is false, but R is true
Answer — (iii)

A is true — the compressions and rarefactions (the disturbance) really do travel through the medium. R is false — the particles do not move forward with the wave; they only oscillate about their mean positions. It is the energy/disturbance that travels, not the particles.

Pause and Ponder • Q5

5. When sound travels from a tuning fork to your ear, which of these 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)

In sound propagation it is the energy that is transferred, not the particles. The air particles only vibrate about their mean positions; the disturbance (energy) travels from the fork to your ear.

Pause and Ponder • Q6

6. Label C and R on the density patterns (Fig. 10.17 a, b), and draw the matching density–distance graphs.

Answer

Mark C (compression) wherever the particles/dots are crowded together (high density) and R (rarefaction) wherever they are spread apart (low density).

In the density–distance graph, each compression becomes a crest (above the average-density dashed line) and each rarefaction becomes a trough (below it). The pattern with more closely spaced C/R gives a wave with shorter wavelength; widely spaced C/R gives a longer wavelength.

Crest (C) Trough (R) Density Distance @edugrown
Compression → crest, Rarefaction → trough (above/below average density)
Pause and Ponder • Q7

7. Repeat Activity 10.1 with a thick and a thin rubber band. Does the thin band vibrate faster? How do its frequency and time period differ?

Answer

Yes, the thin rubber band vibrates faster than the thick one. Faster vibration means a higher frequency (more oscillations per second) and therefore a shorter time period (since $\nu = \tfrac{1}{T}$). The thin band produces a higher-pitched sound.

Pause and Ponder • Q8

8. If a piston produces a sound wave of frequency 20 Hz, how many oscillations does it complete per minute?

Solution
120 Hz means 20 oscillations per second.
2$20 \times 60 = 1200$ oscillations per minute.
1200 oscillations per minute
Pause and Ponder • Q9 • Fig. 10.19

9. For the sound wave shown (markings at 0, 1.5, 3.0, 4.5 cm), what is half of its wavelength?

01.53.04.5 Density Distance (cm) @edugrown
Fig. 10.19 (recreated): one full wave spans 3.0 cm
Solution

One complete wave (crest to next crest) spans from 0 to 3.0 cm, so the wavelength $\lambda = 3.0$ cm.

1Half wavelength $= \dfrac{3.0}{2} = 1.5\ \text{cm}$
Half of the wavelength = 1.5 cm
Pause and Ponder • Q10

10. Compare speeds (Table 10.1): steel 5000, water 1500, air 340 m s⁻¹. Find the ratio of (i) water : air, (ii) steel : water.

Solution
i$\dfrac{v_{water}}{v_{air}} = \dfrac{1500}{340} \approx 4.4$
ii$\dfrac{v_{steel}}{v_{water}} = \dfrac{5000}{1500} \approx 3.3$
Water : Air ≈ 4.4 • Steel : Water ≈ 3.3

So sound travels about 4.4 times faster in water than in air, and about 3.3 times faster in steel than in water.

Pause and Ponder • Q11

11. Two friends are 340 m apart along a steel fence. One knocks; the sound reaches Gunjan through both air and steel. Find the time difference. Can she tell them apart? (Steel 5000 m s⁻¹, air 340 m s⁻¹.)

Solution
1Through air: $t_{air} = \dfrac{340}{340} = 1\ \text{s}$
2Through steel: $t_{steel} = \dfrac{340}{5000} = 0.068\ \text{s}$
3Time difference $= 1 – 0.068 = 0.932\ \text{s}$
Time difference ≈ 0.93 s

Since 0.93 s is much greater than 0.1 s, yes — Gunjan can clearly distinguish the two sounds as separate.

Pause and Ponder • Q12

12. Echoes must arrive at least 0.2 s after the sound. What is the minimum distance of the reflecting surface? (Speed = 343 m s⁻¹.)

Solution
1Total distance (to surface and back) $= v\times t = 343 \times 0.2 = 68.6\ \text{m}$
2Minimum distance $= \dfrac{68.6}{2} = 34.3\ \text{m}$
Minimum distance = 34.3 m
Pause and Ponder • Q13

13. A sonar signal sent to find ocean depth returns in 4 s. If the speed of sound in seawater is 1500 m s⁻¹, find the depth.

Solution
1Depth $= \dfrac{v\times t}{2} = \dfrac{1500 \times 4}{2}$
Depth of ocean = 3000 m
Part 2

Revise, Reflect, Refine — Exercises

Exercise 1

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)

A mechanical wave is one that needs a material medium to travel. Sound cannot travel through vacuum, which directly shows it is mechanical. (Reflection, frequency and energy are shown by many waves, including non-mechanical ones.)

Exercise 2

2. Increasing the frequency of a sound wave in a medium increases its:

(i) wavelength
(ii) speed
(iii) number of compressions per second
(iv) time period
Answer — (iii)

Higher frequency means more density oscillations (compressions) pass per second. The speed stays constant (depends on medium), so the wavelength decreases ($\lambda = v/\nu$) and the time period decreases ($T = 1/\nu$).

Exercise 3

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) 5 Hz
1$\nu = \dfrac{\text{number of oscillations}}{\text{time}} = \dfrac{20}{4} = 5\ \text{Hz}$
Exercise 4

4. Reflected sound reaches the ear 0.05 s after its production. Will it produce an echo or reverberation? Justify.

Answer

It produces reverberation, not an echo. To hear a separate echo, the reflected sound must reach the ear at least 0.1 s after the original. Here the gap is only 0.05 s (less than 0.1 s), so the ear cannot separate the two — the reflected sound merges with and prolongs the original sound, which is reverberation.

Exercise 5 • Fig. 10.30

5. Two sound-wave graphs (same scales). Which has (i) greater wavelength, (ii) smaller amplitude?

Answer

Wave (a) has fewer cycles in the same distance, so it has the greater wavelength. Wave (b) has taller crests/deeper troughs, so it has the larger amplitude — meaning wave (a) has the smaller amplitude.

  • (i) Greater wavelength → wave (a)
  • (ii) Smaller amplitude → wave (a)
Tip: Wavelength = distance between two crests; amplitude = height of a crest above the average line. Compare these directly on your figure.
Exercise 6 • Fig. 10.31

6. Three sources A, B, C. Frequency of A is maximum, C is minimum. Identify and mark the curves.

Answer

Higher frequency = more cycles = shorter wavelength (crests packed closer together). So:

  • A = the curve with the most cycles / closest-spaced crests (highest frequency, shortest wavelength).
  • C = the curve with the fewest cycles / widest-spaced crests (lowest frequency, longest wavelength).
  • B = the curve in between A and C.
Exercise 7

7. Draw a graph for a sound wave with density amplitude 3 units and wavelength 4 cm.

3 units 4 cm (λ) Density Distance @edugrown
Sound wave: amplitude = 3 units, wavelength = 4 cm
Answer

Plot density on the y-axis and distance on the x-axis with the average density as a dashed centre line. Draw a smooth wave whose crests reach +3 units above and troughs go 3 units below the centre line, with each complete cycle (crest to next crest) spanning 4 cm, as shown above.

Exercise 8

8. A movie shows a spacecraft exploding in space with a flash of light and sound at the same time. What are the errors?

Answer
  • Error 1: Space is a vacuum with no medium, so sound cannot travel there — the explosion should be completely silent.
  • Error 2: Even if a medium existed, light travels far faster than sound, so the flash would be seen well before the sound is heard — not at the same instant.
Exercise 9

9. A source produces a sound wave of wavelength 3.44 m travelling at 344 m s⁻¹. Find its time period.

Solution
1Frequency $\nu = \dfrac{v}{\lambda} = \dfrac{344}{3.44} = 100\ \text{Hz}$
2Time period $T = \dfrac{1}{\nu} = \dfrac{1}{100}$
Time period = 0.01 s
Exercise 10

10. A sonar echo is detected after 5 s. If ultrasonic sound travels at 1525 m s⁻¹ in seawater, how deep is the sunken ship?

Solution
1Depth $= \dfrac{v\times t}{2} = \dfrac{1525 \times 5}{2}$
Depth = 3812.5 m
Exercise 11

11. A parking sensor emits ultrasonic waves. When the beep starts, the obstacle is 1.2 m away. How long does the wave take to go to the obstacle and return? (Speed = 345 m s⁻¹.)

Solution
1Total distance $= 2 \times 1.2 = 2.4\ \text{m}$
2$t = \dfrac{\text{distance}}{\text{speed}} = \dfrac{2.4}{345} \approx 0.00696\ \text{s}$
Time ≈ 6.96 × 10⁻³ s (about 7 milliseconds)
Exercise 12

12. Speed of sound is 331 m s⁻¹ at 0 °C and 344 m s⁻¹ at 22 °C. How much extra time does thunder take to travel 1720 m when the temperature changes from 22 °C to 0 °C?

Solution
1At 22 °C: $t_1 = \dfrac{1720}{344} = 5.00\ \text{s}$
2At 0 °C: $t_2 = \dfrac{1720}{331} \approx 5.20\ \text{s}$
3Extra time $= 5.20 – 5.00 \approx 0.20\ \text{s}$
Extra time ≈ 0.2 s
Exercise 13 • Fig. 10.32

13. A sound wave travels at 340 m s⁻¹; the density pattern shows 8 cm spanning two wavelengths. Find the wavelength and frequency.

Solution

From the figure, 8 cm corresponds to 2 wavelengths, so:

1$\lambda = \dfrac{8}{2} = 4\ \text{cm} = 0.04\ \text{m}$
2$\nu = \dfrac{v}{\lambda} = \dfrac{340}{0.04} = 8500\ \text{Hz}$
λ = 4 cm = 0.04 m • ν = 8500 Hz
Note: If your figure shows the 8 cm spanning one wavelength instead, then λ = 8 cm = 0.08 m and ν = 340 ÷ 0.08 = 4250 Hz. Use the spacing exactly as marked in the book.
Exercise 14 • Fig. 10.33

14. Two waves A and B travel at 345 m s⁻¹. From the graph (marks at 0, 2.5, 5.0 cm), find each wavelength and frequency.

A B 02.55.0 Density Distance (cm) @edugrown
Fig. 10.33 (recreated): wave A (shorter λ) and wave B (longer λ)
Solution

From the graph: wave A completes one full cycle in 2.5 cm, wave B in 5.0 cm.

A$\lambda_A = 2.5\ \text{cm} = 0.025\ \text{m},\quad \nu_A = \dfrac{345}{0.025} = 13800\ \text{Hz}$
B$\lambda_B = 5.0\ \text{cm} = 0.05\ \text{m},\quad \nu_B = \dfrac{345}{0.05} = 6900\ \text{Hz}$
A: λ = 2.5 cm, ν = 13800 Hz • B: λ = 5.0 cm, ν = 6900 Hz
Exercise 15 • Fig. 10.34

15. Identical sources at A (in air) and B (in water) send sound to a cliff and back over the same distance. The time to return to A is 4.5 times that to B. Find the ratio of the speeds of sound in air and water.

Solution

Both sounds cover the same distance $d$ (to the cliff and back), and time $t = \dfrac{d}{v}$, so for the same distance time is inversely proportional to speed.

1$t_A = 4.5\,t_B \Rightarrow \dfrac{t_A}{t_B} = 4.5$
2$\dfrac{v_{air}}{v_{water}} = \dfrac{t_B}{t_A} = \dfrac{1}{4.5}$
v(air) : v(water) = 1 : 4.5 (= 2 : 9)

This makes sense — sound travels much faster in water, so it returns to B in much less time.

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