10th Class Physics Unit 13 Sound Numericals with Solutions

13.1.At a particular temperature, the speed of sound in air is 333 m s⁻¹. If the wavelength of a sound wave is 3 cm, calculate the frequency of the sound wave. Is this frequency in the audible range of the human ear?

Given

Speed of sound in air v = 333 m/s

Wavelength of sound wave lambda = 3 cm = 0.03 m

To find

Frequency of the sound wave f = ?

Formula

Using the wave equation v = flambda

Working

Rearranging for frequency f = vlambda

Substituting values f = 3330.03 = 11,100 Hz

Result

The frequency is f = 11.1 kHz (approximately)

Audible range check Audible range: 20 Hz to 20,000 Hz. Since 11.1 kHz lies within this range, it is audible.

13.2.A clock chimes 48 times in 1 minute. Calculate the frequency and period of the chimes.

Given

Number of chimes n = 48

Time taken t = 1 minute = 60 s

To find

Frequency of chimes f = ?

Time period of chimes T = ?

Definition Frequency is the number of oscillations per second

Formula

Frequency f = nt

Working

Substituting values f = 4860 = 0.8 Hz

Definition Time period is the inverse of frequency

Formula

Time period T = 1f

Working

Substituting values T = 10.8 = 1.25 s

f = 0.8 Hz, T = 1.25 s

13.3.A car horn emits a sound with a wavelength of 0.5 m. If the speed of sound in air is 340 m s⁻¹, calculate the frequency of the sound produced by the car's horn.

Given

Wavelength of sound lambda = 0.5 m

Speed of sound in air v = 340 m/s

To find

Frequency of the sound f = ?

Formula

Using the wave equation v = flambda

Working

Rearranging for frequency f = vlambda

Substituting values f = 3400.5 = 680 Hz

f = 680 Hz

13.4.A wave travels with a speed of 500 m s⁻¹ and has a wavelength of 2 metres. Calculate the time period of the wave.

Given

Speed of wave v = 500 m/s

Wavelength lambda = 2 m

To find

Time period of the wave T = ?

Approach First, find frequency using v = flambda

Formula

Frequency from wave equation f = vlambda

Working

Substituting values f = 5002 = 250 Hz

Then calculate time period Time period is the inverse of frequency

Formula

Time period T = 1f

Working

Substituting values T = 1250 = 0.004 s

T = 0.004 s

13.5.A research boat sends a sound wave straight to the seabed and receives the echo 2 seconds later. The speed of sound in seawater is 1600 m s⁻¹. Find the depth of the sea at this position.

Given

Time for echo to return t = 2 s

Speed of sound in seawater v = 1600 m/s

To find

Depth of the sea d = ?

Key insight The sound travels to the seabed and back, so total distance traveled is 2d

Formula

Using distance = speed × time 2d = v × t

Working

Substituting values 2d = 1600 × 2 = 3200 m

Solving for depth d = 32002 = 1600 m

d = 1600 m

13.6.A person claps his hands near a mountain and hears the echo after 6 seconds. If the speed of sound is 343 m s⁻¹, what is the distance of the mountain from the person?

Given

Time for echo to return t = 6 s

Speed of sound in air v = 343 m/s

To find

Distance to the mountain d = ?

Key insight The sound travels to the mountain and back, so total distance traveled is 2d

Formula

Using distance = speed × time 2d = v × t

Working

Substituting values 2d = 343 × 6 = 2058 m

Solving for distance d = 20582 = 1029 m

d = 1029 m

Unit 13: Sound — Numericals