Q1: A source of wave produces 40 crests and 40 troughs in 0.4 seconds. Find the frequency of the wave.
Solution:
Ans: Since there are 40 crests and 40 troughs in 0.4 seconds, there are 40 complete cycles in 0.4 seconds.
As we know that frequency is defined as the number of cycles per second
Hence, To find frequency, we can divide the number of cycles by the time in seconds:
Frequency = (Number of cycles) / (Time in seconds)
Frequency = 40 cycles / 0.4 seconds
Frequency = 100 cycles per second (Hz)
The frequency of the wave is 100 Hz.
Q2. A person has a hearing range from 20 Hz to 20 kHz. What are the typical wavelengths of sound waves in air corresponding to these two frequencies? Take the speed of sound in the air as 344 m s-1
Solution:
Ans: Hearing range = 20 Hz to 20 kHz (= 20,000 Hz) Speed of sound in the air = 344 ms-1 Since, for a wave, Velocity = Wavelength x Frequency
Therefore,
For, v = 20 Hz = 20s-1 λ = 344 / 20 = 17.2 m
For, v = 20,000 Hz = 20,000 s-1 λ = 344 / 20000 = 0.0172 m = 1.72 cm
Q3: Calculate the wavelength of a sound wave whose frequency is 220 Hz and speed is 440 m/s in a given medium.
Solution:
Ans: Given: Frequency, f = 220 Hz, Speed of sound, v = 440 ms-1.
To Find: Wavelegth of sound wave
Since, for a sound wave, Velocity = Wavelength x Frequency ⇒ 440 ms-1 = λ × 220 Hz = λ × 220 s-1 ⇒ λ = 440 / 220 = 2 m Thus, the wavelength of the sound wave is 2 m.
Q4: A person is listening to a sound of 50 Hz sitting at a distance of 450 m from the source of the sound. What is the time interval between successive compressions from the source reaches him?
Solution:
Ans: Frequency of the sound, f = 50 Hz = 50 s-1 , Distance from the source = 450 m.
As we know that, the time between the successive compressions is equal to the time taken by the sound to travel a distance equal to its wavelength.
⇒ Time period, T = 1/Frequency ⇒ T = 1/50 = 0.02 s Thus, the successive compressions will reach the person after every 0.02 s.
Q5: A human heart, on average, is found to beat 75 times a minute. Calculate its frequency.
Solution:
Ans: No. of beats of human heart = 75 min-1. ∵ 1 min = 60 sec. ⇒ No. of beats of human heart = 75/60 = 1.25 s-1 Thus, the average frequency of human heart beating = 1.25 s-1
Q6: A boat at anchor is rocked by waves whose consecutive crests are 100 m apart. The wave velocity of the moving crests is 20 m/s. What is the frequency of rocking of the boat?
Solution:
Ans: Distance between two consecutive crests = wavelength (λ) = 100 m Wave velocity, v = 20 ms-1 The distance between two consecutive crests is equal to the wavelength of the wave(λ). ⇒ Frequency = Velocity/Wavelength. ⇒ f = 20/100 = 0.2 s-1 Thus, the frequency of rocking of the boat is 0.2 s-1
Q7: A longitudinal wave is produced on a toy slinky. The wave travels at a speed of 30 cm/ s and the frequency of the wave is 20 Hz. What is the minimum separation between the consecutive compressions of the slinky
Solution:
Ans: Wave speed, v = 30 cm/ s, Frequency of the wave, f = 20 Hz = 20 s-1.
The minimum separation between the consecutive compressions is equal to the wavelength (λ).
Since , for a wave, Velocity = Frequency ✖ Wavelength
⇒ Therefore, Wavelength, λ = 30/20 = 1.5 cm Thus the minimum separation between the consecutive compression of the slinky is 1.5 cm.
Q8: A bat can hear a sound at frequencies up to 120 kHz. Determine the wavelength of sound in the air at this frequency. Take the speed of sound in the air as 344 m/s.
Solution:
Ans: Frequency, f = 120 kHz = 120 × 103 Hz = 120 × 103 s-1 Velocity of sound in the air, v = 344 ms-1 ∵ Wavelength, λ = Velocity/Frequency ⇒ λ = 344 / (120 x 103) = 2.87 × 10-3 m = 0.29 cm
Q9: A gun is fired in the air at a distance of 660 m, from a person. He hears the sound of the gun after 2s. What is the speed of sound?
Solution:
Ans: Distance travelled by sound = 660 m, Time taken by the sound, T = 2 s. ∵ Speed of sound, v = Distance travelled by sound /time taken by the sound ⇒ v = 660/2 = 330 m/s Thus, The speed of sound in the air is 330 m/s.
Q10: A child hears an echo from a cliff 4 seconds after the sound from a powerful cracker is produced. How far away is the cliff from the child? The velocity of sound in air at 20°C is 344m/s.
Solution:
Ans: Let the distance between the child and the cliff be d. ⇒ Total distance travelled by the sound = 2d Total time taken by the sound, T = 4s,
Take, Velocity of sound, v = 344m/s ⇒ 2d = 344 × 4
d = 688 m Thus, the cliff is at a distance of 688 m from the child.
Q11: A ship sends on a high-frequency sound wave and receives an echo after 1 second. What is the depth of the sea? Speed of sound in water is 1500 m/s.
Solution:
Ans: Let, Depth of the sea = d ⇒ Total distance travelled by the sound wave = 2d Time taken by sound to travel both ways, T = 1s ∵ Speed of the sound = 1500m/s. ⇒ 1500 m/s = 2d/1 ⇒ d = (1500 x 1) / 2 = 750 m Thus, the depth of the sea is 750 meters.
Q12: What are wavelength, frequency, time period and amplitude of a sound wave?
Solution:
Ans:Wavelength: It is the linear distance between two consecutive compressions or two consecutive rarefactions. Frequency: Frequency is the number of complete oscillations (or cycles) per second. Time period : It is the time taken by two consecutive compressions or rarefactions to cross a point. Amplitude: It is the magnitude of maximum displacement of a vibrating particle about its mean position.
Q13: (i) Define the time period of a wave. (ii) Give the relation among speed of sound v, wavelength λ, and its frequency f.
Solution:
Ans: (i) Time period (T) Time period is the time taken for one complete oscillation (cycle).
(ii) Speed of sound (v) = Wavelength (λ) × Frequency (f)
Q14. A wave moves a distance of 8 m in 0.05 s. (a) Find the velocity of the wave. (b) What is the wavelength of the wave if its frequency is 200 Hz?
Solution:
Ans: Distance covered by wave = 8 m, Time taken = 0.05 s. (a) Velocity = 8 / 0.05 = 160 ms-1 (b) λ = v/f = 160/200 = 0.8m
Q15. Two children are at opposite ends of an iron pipe. One strikes his end of the iron pipe with a stone. Find the ratio of times taken by the sound waves in the air and in iron to reach the other child. Given the velocity of sound in air is 344 ms-1 and that in iron is 5130 ms-1.
Solution:
Ans: The distance travelled by sound waves through air and iron is the same, which is equal to the thickness (s) of the rod. Let, The distance travelled = d. For air, let the speed be v1 and time be t1. ⇒ d = v1t1----(i) For iron, let the speed be v2 and time be t2. ⇒ d = v2t2----(ii) From (i) and (ii), ⇒ 14.9.
Q16. A stone is dropped into a well 44.1 m deep. The splash is heard 3.13 seconds after the stone is dropped. Find the velocity of sound in air.
Solution:
Ans: For the downward motion of the stone: s = ut + (1/2) at2
Here, a = acceleration due to gravity, g = 9.8m/s ⇒ 44.1 = (1/2) x 9.8 x t2 ⇒ t = 3 s For reflected sound wave:
Time taken to travel reflected sound wave = 3.13-3 = 0.13 s d = vt ⇒ 44.1 = v x 0.13 ⇒ v = 339.23 ms-1
Q17. A girl sits in the middle of a 12 m × 12 m park. On the left is a building that connects to the park, and on the right of the park is a path that adjoins the park. A sound is created on the road by a cracker. Is it possible for the girl to listen to the echo of this sound? Justify your answer.
Solution:
Ans:An echo will be detected if the gap between the original sound and the reflected sound received at the listener's end is around 0.1 seconds.
Sound velocity × time interval
= 344×0.1
= 34.4 m
where sound reflects from the building and reaches the girl that is way smaller than the desired distance. Hence cannot be heard.
The document Practice Questions with Solutions: Sound is a part of the Class 9 Course Science Class 9.
1. How do sound waves travel through different mediums like air, water, and solids?
Ans. Sound waves travel as mechanical vibrations that require a medium to propagate. In air, sound moves slowest (~340 m/s), while water allows faster transmission (~1480 m/s) due to denser molecular packing. Solids conduct sound fastest because atoms are tightly bonded, enabling rapid vibration transfer. Vacuum blocks sound entirely since no particles exist to carry waves.
2. What's the difference between frequency and pitch in sound, and why does it matter for CBSE exams?
Ans. Frequency measures vibrations per second (measured in Hertz), while pitch is the perceived sensation of high or low sound. Higher frequency produces higher pitch; lower frequency produces lower pitch. This relationship is fundamental to sound wave physics and appears frequently in Class 9 numericals. Understanding this distinction helps solve questions about musical notes and sound characteristics accurately.
3. Can you explain how loudness and amplitude of sound waves are connected?
Ans. Loudness describes the subjective perception of sound intensity, directly dependent on wave amplitude-the maximum displacement of vibrating particles. Greater amplitude means particles vibrate more vigorously, producing louder sound measured in decibels (dB). Doubling amplitude increases loudness noticeably. This amplitude-loudness relationship is essential for answering questions about sound intensity and practical applications in real-world noise measurement.
4. What causes echoes and how is the speed of sound used to calculate distance in echo problems?
Ans. Echoes occur when sound waves reflect off distant surfaces and return to the source. Speed of sound calculations use the formula: Distance = (Speed × Time) ÷ 2, where time is the total duration for sound to travel and return. The division by two accounts for the round trip. This concept appears frequently in Class 9 practice questions involving echo and SONAR applications.
5. Why does sound travel faster in solids than in liquids and gases, and what does this tell us about particle arrangement?
Ans. Sound travels fastest in solids because particles are closely packed and rigidly bonded, enabling efficient vibration transfer through the medium. In liquids, particles have more space but remain relatively close, creating intermediate speed. Gases have maximum particle separation, causing slowest transmission. This particle density relationship explains why underwater sound travels farther than airborne sound, crucial for understanding acoustic behaviour in different environments.
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⇒ 14.9.
