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Class 9 Science | Chapter 10: Sound Waves – Solutions
CHAPTER 10

Sound Waves: Characteristics and Applications

📖 Pages 204–207 | Exploration | Grade 9 Science

🧠 Pause and Ponder pp. 208–215

Q3 (p. 208)
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 option.
✅ Solution
(ii) Both A and R are true, and R is the correct explanation of A. The vacuum bell jar experiment shows that sound cannot travel without a medium.
Q4 (p. 211)
Assertion (A): Compressions and rarefactions move through the medium.
Reason (R): Individual particles of the medium continuously move forward with the wave.
✅ Solution
(iii) A is true, but R is false. Particles only vibrate about their mean positions; the disturbance (compressions/rarefactions) travels, not the particles themselves.
Q7 (p. 213)
Conduct Activity 10.1 with a thick rubber band and then a thin rubber band. Does the thin rubber band vibrate faster? How do frequency and time period differ?
✅ Solution
Yes, the thin rubber band vibrates faster (higher frequency). Higher frequency → shorter time period; lower frequency → longer time period.
Q8 (p. 213)
If frequency of a sound wave is 20 Hz, how many oscillations does the piston complete per minute?
✅ Solution
20 oscillations/second × 60 seconds = 1200 oscillations per minute.
Q9 (p. 213)
For the sound wave represented in the graph (distance between two consecutive crests = 3.0 cm), what is half of its wavelength?
✅ Solution
Wavelength Îģ = 3.0 cm. Half wavelength = Îģ/2 = 1.5 cm.
Q10 (p. 213)
Using Table 10.1 (vwater=1500 m/s, vair=340 m/s, vsteel=5000 m/s), find ratios: (i) speed in water to speed in air, (ii) speed in steel to speed in water.
✅ Solution
(i) 1500/340 = 75/17 ≈ 4.41 → ratio 75:17.
(ii) 5000/1500 = 10/3 ≈ 3.33 → ratio 10:3.
Q11 (p. 213)
Two friends stand 340 m apart along a steel fence. Speed in steel = 5000 m/s, in air = 340 m/s. Calculate time difference. Can they distinguish the two sounds? (Min. distinguishable time = 0.1 s)
✅ Solution
Time through air = 340/340 = 1.0 s.
Time through steel = 340/5000 = 0.068 s.
Difference = 0.932 s (>0.1 s) → Yes, they can hear two separate sounds.
Q12 (p. 215)
An echo requires at least 0.2 s time gap. What minimum distance of reflecting surface? (v = 343 m/s)
✅ Solution
Distance travelled in 0.2 s = 343 × 0.2 = 68.6 m. This is the total to-and-fro path. Minimum distance = 68.6/2 = 34.3 m.
Q13 (p. 215)
A sonar signal returns after 4 s. Speed in seawater = 1500 m/s. Find depth.
✅ Solution
Total distance = 1500 × 4 = 6000 m. Depth = half = 3000 m.

📘 Revise, Reflect, Refine pp. 204–207

Q1
Which observation best supports that sound is a mechanical wave? (i) Reflection (ii) Needs a medium (iii) Has frequency (iv) Carries energy
✅ Answer
(ii) Sound needs a medium to propagate.
Q2
For a sound wave, 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 = number of compressions per second).
Q3
If 20 compressions pass a point in 4 seconds, find frequency.
✅ Solution
20 compressions = 20 complete waves. Frequency = 20/4 = 5 Hz.
Q4
In a room, reflected sound reaches the ear 0.05 s after production. Will it produce an echo or reverberation?
✅ Answer
Reverberation (time gap < 0.1 s, so multiple reflections blend with original sound).
Q5
Two sound wave graphs with same axes. Which has (i) greater wavelength, (ii) smaller amplitude?
✅ Answer
(i) The wave with larger distance between consecutive crests (longer Îģ).
(ii) The wave with smaller peak height (lower amplitude).
Q6
Three curves (green, red, blue). Frequency of A is max, C is min. Identify A, B, C.
✅ Answer
A (highest frequency) → most oscillations (green).
C (lowest frequency) → fewest oscillations (blue).
B → medium (red).
Q7
Draw a graph for sound wave with density amplitude = 3 units, wavelength = 4 cm.
✅ Description
A sine-like curve with vertical axis showing density variation (amplitude 3) and horizontal axis distance. Mark crest-to-crest distance = 4 cm.
Q8
In a movie, explosion in space shown with simultaneous flash and sound. What errors?
✅ Answer
Space is near vacuum; sound cannot travel, so no explosion sound should be heard – only light (flash).
Q9
Wavelength = 3.44 m, speed = 344 m/s. Find time period.
✅ Solution
f = v/Îģ = 344/3.44 = 100 Hz. T = 1/f = 0.01 s.
Q10
Sonar echo after 5 s, speed in seawater = 1525 m/s. Find depth.
✅ Solution
Total distance = 1525 × 5 = 7625 m. Depth = 7625/2 = 3812.5 m.
Q11
Distance to obstacle = 1.2 m, speed = 345 m/s. Time for wave to go and return.
✅ Solution
Total distance = 2.4 m. Time = 2.4 / 345 ≈ 0.00696 s ≈ 0.007 s.
Q12
Distance = 1720 m. Speed at 22°C = 344 m/s, at 0°C = 331 m/s. Find extra time at 0°C.
✅ Solution
t22 = 1720/344 = 5 s. t0 = 1720/331 ≈ 5.196 s. Extra time ≈ 0.196 s.
Q13
Graph shows distance between two compressions = 8 cm. Speed = 340 m/s. Find Îģ and f.
✅ Solution
Îģ = 8 cm = 0.08 m. f = 340 / 0.08 = 4250 Hz.
Q14
Two waves A and B with same speed = 345 m/s. ÎģA = 2.5 cm, ÎģB = 5.0 cm. Find frequencies.
✅ Solution
fA = 345 / 0.025 = 13800 Hz. fB = 345 / 0.050 = 6900 Hz.
Q15
Two identical sources A (air) and B (water). Time to return to A is 4.5 times that to B. Find vair/vwater.
✅ Solution
Let distance = d. tA = 2d/vair, tB = 2d/vwater. Given tA = 4.5 tB → 1/vair = 4.5/vwater → vair/vwater = 1/4.5 = 2/9. Ratio = 2:9.

📐 Worked Examples (from chapter) pp. 196–203

Example 10.1
If there are 10 density oscillations in 2 seconds at a given position, calculate (i) frequency, (ii) time period.
✅ Solution
f = 10/2 = 5 Hz. T = 1/f = 0.2 s.
Example 10.2
Human hearing 20 Hz to 20 kHz. Speed of sound = 344 m/s. Find corresponding wavelengths.
✅ Solution
Îģ20Hz = 344/20 = 17.2 m. Îģ20kHz = 344/20000 = 0.0172 m = 1.72 cm.
Example 10.3
Time delay between lightning and thunder = 5 s, v = 340 m/s. Estimate distance to lightning.
✅ Solution
Distance = 340 × 5 = 1700 m (1.7 km).
Example 10.4
Graph of sound wave in steel, Îģ = 50 m, v = 5000 m/s. Find frequency and time period.
✅ Solution
f = 5000/50 = 100 Hz. T = 1/100 = 0.01 s.
Example 10.5
Clap in empty corridor, echo after 0.5 s, v = 340 m/s. Find distance from wall.
✅ Solution
Distance = (340 × 0.5)/2 = 85 m.
Example 10.6
Sonar signal returns after 0.90 s, vseawater = 1530 m/s. Find distance to object.
✅ Solution
Distance = (1530 × 0.90)/2 = 688.5 m.

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📚 Exploration | Grade 9 Science | NCERT 2026-27
Chapter 10: Sound Waves – Characteristics and Applications (Complete Solutions)

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