Numericals with Answers - Sound | Science Class 9 PDF Download

The study of Class 9 Science Chapter - Sound is an exciting part of your Science curriculum. Sound is all around us, from the chirping of birds to the music we enjoy. In this chapter, you’ll learn how sound is produced through vibrations, how it travels as a wave through different mediums like air, water, or solids, and how we hear it.

 You’ll explore key concepts like the speed of sound, its frequency, wavelength, and the fascinating phenomenon of echoes. Understanding sound will help you appreciate its role in communication, technology, and even nature. 

Basic Formulae

Speed of Sound: The speed of sound (v) is related to its frequency (f) and wavelength (λ).

v = f × λ

Unit: Speed in m/s, frequency in Hz, wavelength in m.

Frequency and Time Period: Frequency (f) is the number of oscillations per second, and time period (T) is the time for one oscillation.

f = 1 / T

T = 1 / f

Unit: Frequency in Hz, time period in seconds.

Distance for Echo: To hear a distinct echo, the time gap between the original sound and the echo must be at least 0.1 seconds.

Total distance traveled by sound = Speed of sound × Time

Distance to reflecting surface = (Speed of sound × Time) / 2 (since sound travels to the surface and back).

Speed of Sound in Different Media: The speed varies depending on the medium (e.g., 331.45 m/s in air, 1496 m/s in water at a given temperature) and increases with temperature.

Solved Examples

Sol: Formula: v = f × λ

Given: f = 500 Hz, λ = 0.7 m

v = 500 × 0.7 = 350 m/s

Q2: A sound wave has a time period of 0.02 seconds. Calculate its frequency.

Sol: Formula: f = 1 / T

Given: T = 0.02 s

f = 1 / 0.02 = 50 Hz

Q3: The speed of sound in air is 340 m/s, and the frequency of a sound wave is 400 Hz. Calculate the wavelength of the sound wave.

Sol: Formula: v = f × λ

Given: v = 340 m/s, f = 400 Hz

λ = v / f = 340 / 400 = 0.85 m

Q4: A boy claps near a wall and hears an echo after 0.4 seconds. If the speed of sound is 340 m/s, calculate the distance between the boy and the wall.

Sol: Total distance traveled by sound = Speed × Time

Given: Speed = 340 m/s, Time = 0.4 s

Total distance = 340 × 0.4 = 136 m

Distance to the wall = Total distance / 2 = 136 / 2 = 68 m

Q5: A girl shouts near a cliff and hears an echo after 3 seconds. If the speed of sound is 342 m/s, calculate the distance between the girl and the cliff.

Sol:Total distance traveled by sound = Speed × Time

Given: Speed = 342 m/s, Time = 3 s

Total distance = 342 × 3 = 1026 m

Distance to the cliff = Total distance / 2 = 1026 / 2 = 513 m

Q6: What is the minimum distance required to hear a distinct echo if the speed of sound in air is 345 m/s? (Assume the minimum time for a distinct echo is 0.1 seconds.)

Sol: Total distance traveled by sound = Speed × Time

Given: Speed = 345 m/s, Time = 0.1 s

Total distance = 345 × 0.1 = 34.5 m

Minimum distance to the reflecting surface = Total distance / 2 = 34.5 / 2 = 17.25 m

Q7: A sound wave has a frequency of 1000 Hz. Calculate its wavelength in air (speed = 340 m/s) and in water (speed = 1496 m/s). Compare the wavelengths.

Sol: Formula: λ = v / f

In air: v = 340 m/s, f = 1000 Hz

λ_air = 340 / 1000 = 0.34 m

In water: v = 1496 m/s, f = 1000 Hz

λ_water = 1496 / 1000 = 1.496 m

Comparison: Ratio of wavelengths = λ_water / λ_air = 1.496 / 0.34 ≈ 4.4

⇒ Ratio ≈ 4.4:1

Q8: A sound wave travels in air at 336 m/s with a wavelength of 0.8 m. Calculate its frequency and time period.

Sol: Frequency:

Formula: v = f × λ

Given: v = 336 m/s, λ = 0.8 m

f = v / λ = 336 / 0.8 = 420 Hz

Time period:

Formula: T = 1 / f

T = 1 / 420 ≈ 0.00238 s