NCERT Solutions for Class 9 Science Chapter 12 Sound

Q. What are wavelength, frequency, time period and amplitude of a sound wave ?

Ans. Wavelength: Wavelength is the distance between two consecutive rarefactions or two consecutive compressions. The S.I. unit of wavelength is metre (m).
Frequency: Frequency is the number of oscillations per second. The S.I. unit of frequency is hertz (Hz).
Time period: The time period is the time required to produce one complete cycle of a sound wave.
Amplitude: Amplitude is the maximum height reached by the trough or crest of a sound wave.

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

Ans. Given, Frequency of sound wave = 220 Hz
Speed of sound wa
ve = 440 m/s
To calculate wavelength,
As speed is giv
en by,
Speed = Wavelength × Frequency
v = λn
440 = Wavelength × 220
Wavelength =440/220
Wavelength = 2m
Thus, the wavelength of the sound wave = 2 metres.

Q. An echo is heard in 3s. What is the distance of the reflecting surface from the source, given that the speed of sound is 342 m/s?

Ans. Given, Speed of sound (v) = 342 m/s Echo returns in time (t) = 3 s
Distance travelled by sound waves = v × t
= 342 × 3
= 1026 m
In the interval of time, sound must travel twice the distance of reflecting surface to source.
Thus, the distance of reflecting surface from the source
= 513 m

Q. Flash and thunder are produced simultaneously. But thunder is heard a few seconds after the flash is seen, why?

Ans. The speed of sound is around 344 m/s whereas the speed of light is 3 × 108 m/s. The speed of sound is less when compared to light. Due to this reason, the thunder sound takes more time to reach the Earth as compared to the light speed which is faster. Hence, lightening is seen before we hear the thunder.

Q. How is ultrasound used for cleaning?

Ans. Objects to be cleaned are kept in a cleaning solution and ultrasonic sound waves are passed through the solution. The high frequency of ultrasound waves helps in detaching the dirt from the objects. By this way ultrasound is used for cleaning purposes.

Q. Two children are at opposite ends of an aluminium rod. One strikes the end of the rod with a stone. Find the ratio of times taken by the sound wave in the air and in aluminium to reach the second child.

Ans. Let length of aluminium rod = d Speed of sound wave at 25°C, vAl = 6420 m/s in aluminium Time tak en by sound through aluminium rod

$$t_{Al}=\frac{d}{(V_{al})}\\ =\frac{d}{6420}\\ \text{Speed of sound in air,}\\ v_{air} = 346 ms^{–1}\\ \text{Time taken by sound through air}\\ t_{air}=\frac{d}{(V_{al})}=\frac{d}{346}\\ \text{Thus, the ratio of time taken by sound in aluminium and air.}\\ \frac{t_{air}}{t_{Al}}=\frac{6420}{346}=18.55$$

Q. A stone is dropped from the top of a tower of height 500 m into a pond of water at the base of the tower. When is the splash heard at the top? Given, g = 10m/s2 and speed of sound = 340 m/s.
Ans. Given, Height of tower = 500 m Velocity (v) of sound = 340 m/s Acceleration (g) due to gravity = 10 m/s Initial velocity (u) of the stone = 0 Time(t1) taken by the stone to fall to tower based (pond of water) from the equation of motion,

$$s =ut_1+\frac{1}{2}g (t_1)^2\\ 500 =0 × t_1+\frac{1}{2}10 (t_1)^2\\ (t_1)^2 = 100\\ t_1 = 10 s\\ Time (t_2) \text{taken by sound to reach top from tower base}\\ =\frac{Height of tower}{Speed of sound in air}\\ t_2 =\frac{500}{340}=1.47s\\ t = t_1 + t2\\ t = 10 + 1.47\\ t = 11.47 sec$$

Q. A sound wave travels at a speed of 339 m/s. If its wavelength is 1.5 cm, what is the frequency of the wave? Will it be audible?

Ans. Given, Speed (v) of sound = 339 m/s
Wavelength (λ) of sound = 1.5 cm = 0.015 m
Speed of sound = Wavelength × Frequency
v = λ × n
n =v/ λ
= 339/0.015
= = 22600 Hz

The frequency of audible sound for human beings lies between the ranges of 20 Hz to 20,000 Hz. The frequency of the given sound is more than 20,000 Hz, therefore, it is not audible range of human, it is ultrasonic sound audiable by Bats.