Class 9 Science Chapter 11 Practice Question Answers - Sound

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.

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

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.

Ans: Frequency of the sound, f = 50 Hz = 50 s^-1 , Distance from the source, λ = 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.

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

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   

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.

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

Ans: Distance travelled by sound, λ = 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.

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.

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.

Ans: Wavelength: It is the linear distance between two consecutive compressions or two consecutive rarefactions.
Frequency: The number of compressions or rarefactions taken together passing through a point in one second is called frequency.
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.

Ans: (i) Time period (T)
It is defined as the time required to complete one wave.

(ii) Speed of sound ($v$v) = Wavelength ($\backslash lambda$λ) × Frequency ($f$f)

Ans: Distance covered by wave = 8 m, Time taken = 0.05 s.
(a) Velocity = 8 / 0.05 = 160 ms^-1
(b) λ = v/f = 0.8m

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, jet 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), v₁t₁ = v₂t₂
⇒ t₁/t₂ = 5130 / 344 = 14.9

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

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 can not be heard.