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The human ears are able to hear sound in a frequency range of about 20 Hz to 20 kHz. Sounds of frequency less than 20 Hz are known as infrasonic sounds.

Sound of loudness lower than 80 dB can be tolerated by a normal human being.

a) Higher than 20,000 Hz

Sound is produced by Vibrating objects only.

Unit of Intensity of loudness is (dB) decibels

Sound cannot travel through Vacuum.

Normal range of loudness is 50 (dB) to 80 (dB).

Vibration is also known as Oscillatory motion.

D) Both A & B

Correct option is C) 1484 m/s

Transverse waves

Correct option is B) transverse

Correct Answer is: (c, d)

Correct option is A) v = λυ

a) Longitudinal waves 

b) Transverse waves

D) Both A & B

Correct option is B) hertz

Speed, wavelength, and frequency of a sound wave are related by the following equation:

Speed (v) = Wavelength (λ) × Frequency (v)

v = λ x v

Like all waves, sound waves can be reflected. Sound waves suffer reflection from the large obstacles. As a result of reflection of sound wave from a large obstacle, the sound is heard which is named as an echo. Ordinarily echo is not heard as the reflected sound gets merged with the original sound. Certain conditions have to be satisfied to hear an echo distinctly (as a separate sound). The sensation of any sound persists in our ear for about 0.1 seconds. This is known as the persistence of hearing. If the echo is heard within this time interval, the original sound and its echo cannot be distinguished. So the most important condition for hearing an echo is that the reflected sound should reach the ear only after a lapse of at least 0.1 second after the original sound dies off. As the speed of sound is 340 m/s, the distance travelled by sound in 0.1 second is 34 m. This is twice the minimum distance between a source of sound and the reflector. So, if the obstacle is at a distance of 17 m at least, the reflected sound or the echo is heard after 0.1 second, distinctly.

Further, for reflection of any wave to take place, the size of the reflector should be large compared to the wavelength of the sound, which for ordinary sound is of the order of 1 metre. A large building, a mountain side, large rock formation etc. are good reflectors of sound for producing an echo. Also, for the reflected sound to be heard, it must have enough intensity or loudness. Moreover, if the echo is to be distinguished from the original sound the two should not mix or overlap. For this, the original sound should be of very short duration, like a clap or shout.

So, following conditions could be listed for formation of echo:

• The size of the obstacle/reflector must be large compared to the wavelength of the incident sound (for reflection of sound to take place).

• The distance between the source of sound and the reflector should be at least 17 m (so that the echo is heard distinctly after the original sound is over).

• The intensity or loudness of the sound should be sufficient for the reflected sound reaching the ear to be audible. The original sound should be of short duration.

Correct Answer is: (a) I_max = 0.64 I₀ , (c) I_max /I_min = 16

Let a = initial amplitude due to S₁ and S₂ each.

I₀ = k(4a²), where k is a constant.

After reduction of power of S₁, amplitude due to S₁ = 0.6a. 

Due to superposition, a_max = a + 0.6a = 1.6a, and

a_min = a - 0.6a = 0.4a

I_max/I_min = (a_max/a_min)² = (1.6a/0.4a) = 16.

Correct Answer is: (d) the overtones present in the sound

The phenomenon of propagation of original sound due to the multiple reflections of sound waves even after the source of sound and stops producing sound is called reverberation. It can be reduced by covering the roofs and walls of the hall by absorbing materials.

Persistence of sound (after the source stops producing sound) due to repeated reflection is known as reverberation. As the source produces sound, it starts travelling in all directions. Once it reaches the wall of a room, it is partly reflected back from the wall. This reflected sound reaches the other wall and again gets reflected partly. Due to this, sound can be heard even after the source has ceased to produce sound.

To reduce reverberations, sound must be absorbed as it reaches the walls and the ceiling of a room. Sound absorbing materials like fibreboard, rough plastic, heavy curtains, and cushioned seats can be used to reduce reverberation.

Correct option is A) music

Unwanted sounds are called  noise

The following measures which can be taken to control (or reduce) noise pollution in our surroundings:

1. Ban the use of loudspeakers after 9 pm. 

2. Ban the use of horns in the areas nearby hospital and school.

Correct option is D) pollution

Correct option is B) noise

Frequency of oscillations is the number of oscillations of a vibrating object per second. Therefore frequency is = 40 vibrations /4 seconds= 10 Hertz. 

Time period is the time required to complete one oscillation. Or it is the inverse of time period. 

Therefore time period = 1/10 = 0.1 seconds.

Correct option is D) B and C

It is very unpleasant to hear.

Correct option is B) 200 Hz

No, students cannot study in this condition. They cannot concentrate on their study.

Correct option is  C) Producing sound

This atmosphere is not convenient for a patient who is ill. He cannot take rest.

1. Chirping of bird outside the house on the tree.

2. Collision of two vehicles – honking of the vehicles.

3. Baby crying.

4. Father talking on the phone.

5. Brother playing in the house – jumping.

6. Sister playing the drum – hitting the drum.

7. Sound of TV.

8. Dog barking.

. Mother talking to the neighbour.

High pitch = 512 Hz

Low pitch = 256 Hz

Songs from radio are more pleasant to hear.

i. The singing of the girl is a pleasant sound.

ii. 1. Boy shouting/screaming.

2. Boy moving/ walking with a toy rattle.

i. Nature of the object 

ii. Length 

iii. Tention 

iv. Surface area 

v. Area of cross-section

Given 

Speed of sound (u) = 334 m/s 

Time taken for hearing the echo (t) = 1.5s

Now 

speed = \(\frac{distance}{time}\)

⇒334 = \(\frac{distance}{1.5}\) 

⇒ Distance travelled by the sound = 334 ×1.5 = 501 m.

In 1.5 s, sound has to travel twice the distance between the source and surface. Therefore, the distance between the source and surface is =\(\frac{501}{2}\) = 250.5 m. 

Distance of source from surface = 250.5 m

Frequency = 100Hz.
From the definition of frequency, we can say that Number of oscillations in 1 min or 60 Sec. = 100 × 60
= 6000 = 6 × 10³.

Yes. Sound waves follow the same laws of reflection as light does. The directions in which the sound is incident and is reflected make equal angles with the normal to the reflecting surface at the point of incidence and three are in the same plane.

The time taken by echo to be heard

t = \(\frac{2d}{v}\)

where d = distance between the reflecting surface and source of sound and v = speed of sound in air. As we know that speed of sound increases with increase in temperature. So on a hotter day speed of sound will be higher, so the time after which echo is heard will decrease.

If time taken by the reflected sound is less than 0.15 after the production of original sound, then echo is not heard.