NCERT Class 9 Science Sound Waves Characteristics and Applications Solutions
Short Intro
This chapter explains how sound is produced, propagated, reflected, and perceived by humans. Students will learn about sound waves, compressions and rarefactions, wavelength, frequency, amplitude, echo, reverberation, ultrasonic waves, sonar, and practical applications of sound in daily life.
Quick Information Box
| Topic | Details |
|---|---|
| Chapter Name | Sound Waves: Characteristics and Applications |
| Class | Grade 9 |
| Subject | Science |
| Wave Type | Longitudinal Mechanical Wave |
| Audible Range | 20 Hz to 20 kHz |
| Speed of Sound in Air | Approximately 340 m/s |
Concepts Used (Topics Covered)
- Production of Sound
- Propagation of Sound
- Longitudinal Waves
- Compression and Rarefaction
- Mechanical Waves
- Wavelength, Frequency and Time Period
- Amplitude and Loudness
- Speed of Sound
- Echo and Reverberation
- Reflection of Sound
- Ultrasonic and Infrasonic Waves
- Sonar and Echolocation
Important Formulas
- Frequency
ν = Number of Oscillations / Time - Time Period
T = 1 / ν - Speed of Sound
v = νλ - Distance Formula
Distance = Speed × Time - Echo Distance
Distance = (v × t)/2
Questions & Step-by-step Solutions with Explanation
Q1. Sound as Mechanical Wave
Question:
Which observation best supports that sound is a mechanical wave?
Options:
(i) Sound shows reflection
(ii) Sound needs a medium to propagate
(iii) Sound has frequency
(iv) Sound carries energy
Solution:
Mechanical waves require a material medium for propagation.
Answer:
(ii) Sound needs a medium to propagate
Q2. Frequency and Oscillations
Question:
If 20 compressions pass a point in 4 seconds, find frequency.
Formula:
Frequency = Oscillations / Time
Calculation:
ν = 20 / 4
ν = 5 Hz
Answer:
5 Hz
Q3. Echo or Reverberation
Question:
Reflected sound reaches ear after 0.05 s. Is it echo or reverberation?
Solution:
Echo requires minimum 0.1 s gap.
Since time gap is less than 0.1 s, reflected sound overlaps with original sound.
Answer:
It is reverberation.
Q4. Compare Sound Waves
Solution:
- Greater wavelength → wave with larger crest-to-crest distance
- Smaller amplitude → wave with smaller height
Students should identify directly from graph.
Q5. Frequency Identification
Solution:
Higher frequency means more oscillations in same distance.
- A → Highest frequency
- C → Lowest frequency
- B → Intermediate frequency
Q6. Draw Sound Wave
Solution:
Wave should have:
- Amplitude = 3 units
- Wavelength = 4 cm
Draw proper crest and trough distances.
Q7. Spacecraft Explosion Error
Solution:
Two errors:
- Sound cannot travel in vacuum of space.
- Sound and light cannot arrive together because sound is much slower.
Q8. Time Period Calculation
Given:
Wavelength = 3.44 m
Speed = 344 m/s
Formula:
v = νλ
ν = v/λ
ν = 344 / 3.44
ν = 100 Hz
Time Period:
T = 1/ν
T = 1/100
T = 0.01 s
Answer:
0.01 second
Q9. Sonar Problem
Given:
Time = 5 s
Speed = 1525 m/s
Formula:
Distance = (v × t)/2
Calculation:
Distance = (1525 × 5)/2
= 3812.5 m
Answer:
Depth of ocean = 3812.5 m
Q10. Ultrasonic Parking Sensor
Given:
Distance = 1.2 m
Speed = 345 m/s
Total Distance:
= 2 × 1.2
= 2.4 m
Time:
t = Distance / Speed
= 2.4 / 345
≈ 0.00696 s
Answer:
0.007 second approximately
Q11. Thunder Sound Delay
Given:
Distance = 1720 m
At 22°C:
Time = 1720 / 344
= 5 s
At 0°C:
Time = 1720 / 331
≈ 5.2 s
Extra Time:
= 5.2 − 5
= 0.2 s
Answer:
Thunder takes approximately 0.2 s extra.
Q12. Frequency from Graph
Given:
Wave speed = 340 m/s
Wavelength from graph = 8 cm
Convert:
8 cm = 0.08 m
Formula:
ν = v/λ
ν = 340 / 0.08
= 4250 Hz
Answer:
Frequency = 4250 Hz
Q13. Frequency of Waves A and B
Given:
Speed = 345 m/s
Students measure wavelength directly from graph.
Formula:
ν = v/λ
Use wavelength values from graph for calculation.
Q14. Sound in Air and Water
Given:
Time in air = 4.5 × Time in water
Since distance is same:
Speed ∝ 1/Time
Ratio:
Speed in air : Speed in water
= 1 : 4.5
Simplified:
2 : 9
Answer:
2 : 9
Q15. Assertion and Reason
Assertion:
We cannot hear bell ringing in evacuated jar.
Reason:
Sound requires medium to travel.
Answer:
(ii) Both A and R are true and R correctly explains A.
Q16. Compression and Rarefaction Assertion
Assertion:
Compressions and rarefactions move through medium.
Reason:
Particles continuously move forward with wave.
Solution:
Particles only vibrate about mean positions.
Answer:
(iii) A is true but R is false.
Q17. Energy Transfer in Sound
Question:
What actually reaches ear from tuning fork?
Solution:
Energy travels through sound waves.
Particles themselves do not travel to ear.
Answer:
(ii) Energy carried by sound waves
Common Mistakes
- Confusing loudness with intensity
- Using wrong units for wavelength
- Forgetting echo distance is half distance travelled
- Mixing frequency and time period
- Assuming particles travel with sound wave
Exam Tips
- Learn all sound formulas carefully.
- Practice numerical problems regularly.
- Remember audible range values.
- Draw labelled wave diagrams neatly.
- Always convert cm into m before calculations.
Practice MCQs
1. Sound waves are:
A. Transverse waves
B. Longitudinal waves
C. Electromagnetic waves
D. Stationary waves
Answer:
B. Longitudinal waves
2. SI unit of frequency:
A. Joule
B. Newton
C. Hertz
D. Pascal
Answer:
C. Hertz
3. Human audible range:
A. 2 Hz – 200 Hz
B. 20 Hz – 20 kHz
C. 200 Hz – 2 MHz
D. 100 Hz – 1000 Hz
Answer:
B. 20 Hz – 20 kHz
4. Sound cannot travel through:
A. Air
B. Water
C. Steel
D. Vacuum
Answer:
D. Vacuum
5. Instrument used in underwater detection:
A. Radar
B. Sonar
C. Microscope
D. Periscope
Answer:
B. Sonar
FAQ Section
Q1. Why is sound called a mechanical wave?
Because it requires a material medium to travel.
Q2. What is wavelength?
Distance between two consecutive compressions or rarefactions.
Q3. What is frequency?
Number of oscillations per second.
Q4. Why do we hear echo?
Due to reflection of sound from distant surfaces.
Q5. What are ultrasonic waves?
Sound waves having frequency above 20 kHz.
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