# Sound class 9 Numerical with solutions

In this post, we will solve a set of **Numerical problems from the Sound chapter of class 9 science**.

**Sound Class 9 Numerical Questions and Answers** [solved]

**Sound Class 9 Numerical Questions and Answers**[solved]

**1) A sound wave has a frequency of 2 kHz and a wavelength of 35 cm. How long will it take to travel 1.5 km?**

**Solution:**

Given,

Frequency, n = 2 kHz = 2000 Hz

Wavelength, l = 35 cm = 0.35 m

We know that speed, v of the wave = wavelength × frequency

=> v = l n

=> v = 0.35 m × 2000 Hz = 700 m/sThe time taken by the wave to travel a distance, d of 1.5 km is t = d/v = (1.5×1000) m/700 m/s =2.1 s

Thus sound will take 2.1 seconds to travel a distance of 1.5 km.

**2 ) Calculate the wavelength of a sound wave whose frequency is 220 Hz and the speed is 440 m/s in a given medium.**

Solution:

speed of the wave = wavelength × frequency

=> wavelength = speed of the wave/frequency = 440/220 m = 2 m

**3 ) A person is listening to a tone of 500 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?**

Solution:

Given: frequency = 500 Hz

The time interval between successive compressions is equal to the time period of the wave.

This time period is the reciprocal of the frequency of the wave.

It is given by: T = 1 / Frequency = 1 / 500 s=0.002 s

**4 ) A person clapped his hands near a cliff and heard an echo after 2 seconds.What is the distance of the cliff from the person if the speed of the sound, v is taken as 346 m/s?**

Solution:

Given,

Speed of sound, v = 346 m/s

Time taken to hear the echo, t = 2 s

Distance traveled by the sound = v × t = 346 m/s × 2 s = 692 m

In 2 s sound has to travel twice the distance between the cliff and the person. Hence, the distance between

the cliff and the person = 692 m/2 = 346 m.

**5 ) A ship sends out an ultrasound that returns from the seabed and is detected after 3.42 s. If the speed of ultrasound through seawater is 1531 m/s, what is the distance of the seabed from the ship?**

Solution:

Given,

The time between transmission and detection, t = 3.42 s.Speed of ultrasound in seawater, v = 1531 m/s

Distance traveled by the ultrasound = 2 × depth of the sea = 2d

where d is the depth of the sea.2d = speed of sound × time

=> 2d = 1531 m/s × 3.42 s = 5236 m

d = 5236 m/2 = 2618 m.

Thus, the distance of the seabed from the ship is 2618 m or 2.62 km.

**6 ) A submarine emits a sonar pulse, which returns from an underwater cliff in 1.02 s. If the speed of sound in salt water is 1531 m/s, how far away is the cliff?**

Solution:

Given,

The time between transmission and detection, t = 1.02 s.Speed of sonar in salt water, v = 1531 m/s

Distance traveled by the ultrasound = 2 × depth of the cliff = 2d

where d is the depth of the underwater cliff.2d = speed of sound × time

=> 2d = 1531 m/s × 1.02 s = 1562 m

d = 1562 m/2 = 781 m.

Thus, the distance of the cliff from the submarine is 781 m.

**7 ) An echo is heard after 0.8 s when a boy fires a cracker, 132m away from a tall building. Calculate the speed of sound. **

Solution:

Echo time (t) = 0.8s

Total distance traveled by sound wave = 2d = 2×132 m = 264 m

From, speed of sound V = 2d/t

=> V =264/0.8 m/s = 330 m/s