Sound

• When the school bell vibrates, it forces the adjacent particles in air to vibrate.
• This disturbance gives rise to a wave and when the bell moves forward, it pushes the air in front of it.
• This creates a region of high pressures known as compression.
• When the bell moves backward, it creates a region of low pressure know as rarefaction.
• As the bell continues to move forward and backward, it produces a series of compressions and rarefactions. T
• his makes the sound of a bell propagate through air.

No, Since sound waves are longitudinal waves which require medium to propagate in vaccum which has no particles to vibrate. So, I will not able to hear the sound of my friend on the moon.

• Sound intensity is a property of the sound source but loudness depends on the sound source, the medium and the receiver, as well.
• Sound intensity holds a small significance in problems involving human hearing system, but loudness is a very important property to consider in such problems.
• Sound intensity is measured in Watt per square meter whereas loudness is measured in Sones.

Wavelength: Wavelength is the distance between two consecutive compressions or rarefaction of wave.

Frequency: The number of sound wave produced in one second is called frequency.

Time period: Time period is the time taken to produce one wave of sound.

Amplitude: Amplitude is the maximum displacement along the mean position of the particles of medium.

(a) Amplitude of sound waves determines loudness. Louder sound has greater amplitude and vice versa.

(b) Frequency of the sound waves determines pitch of the sound.

The relation between frequency and wavelength of sound wave is given as follows:

Velocity (v) = Wavelength (λ) * Frequency (ν)

v = λ * ν

This means the speed is equal to the product of wavelength and frequency of the sound wave.

This equation is also called the ‘wave equation’ and applicable to all types of wave.

The sound of the guitar has higher pitch since, The frequency of the wave determines pitch, i.e., higher the frequency of a wave, more is its pitch. The frequency of vibration of the particle is more in case of the guitar than car horn. Hence, guitar has higher pitch than car horn

We know that,

v = f × λ (i)

Given,

Speed, v = 440 m/s

Frequency, f = 220 Hz

Wavelength λ =?

Putting these values in (i), we get:

440 = 220 × λ

λ =

= 2 m

Given,

Frequency, f = 500 Hz

Time period, T =

=

= 0.002 s

Therefore, the time interval between successive compressions from the source of sound is equal to the time-period of sound waves.

At particular temperature sound travels fastest in iron as we know that sound travels fastest in any solid medium and iron is solid.

The time taken by sound to travel from the source to the reflecting surface will be half of this time =

= 1.5 seconds

We know that,

Speed =

342 =

Distance = 342 × 1.5

= 513 m

Since, concert halls are big, so audience at the back rows of the hall may not hear clear sound of speaker. To overcome this problem, the ceiling of the concert halls is made concave. Concave ceiling helps the sound wave to reflect and send to farther distance which makes the concert hall enable to send clear sound to the audience even sitting in back rows of hall.

The range of frequencies of a human ear is from 20 Hz to 20,000 Hz.

(a) Range of frequencies with infrasonic sound is less than 20 Hz.

(b) Ultrasonic frequencies range higher than 20000Hz.

The time taken by the sonar pulse to go from the submarine to cliff will be half of this time:

=

= 0.51 second

We know that,

Speed =

1531 =

Distance = 1531 × 0.51

= 780.8 m

Since sound wave creates oscillation in the particles of the medium parallel to the disturbance in the direction of propagation, thus sound waves are called longitudinal wave. This would be more clear by taking the definition of longitudinal wave into account.

Longitudinal wave: When oscillation is created parallel to the disturbance of the particles of medium in the direction of propagation.

Timbre and pitch are the characteristics of sound which help to identify the sound of different voice. Thus, because of difference in timbre and pitch of the sound wave I or any other can identify the voice of his friend sitting with others even in dark room.

This happens because of the difference in the velocity of light and sound waves. Light travels with much faster velocity than sound. That’s why thunder is heard a few seconds after the flash of thunder is seen instead of both are produced simultaneously.

Given,

Speed, v = 344 m/s

Frequency, f = 20 Hz

Wavelength, λ₁ =?

We know that,

v = f × λ

Putting the values in the above equation, we get

λ₁= = 17.2 m

When the frequency is 2 KHz:

Wavelength, λ₂=?

λ₂= = 0.0172

Therefore, the wavelength of sound in air having frequencies corresponding 20 Hz is 17.2 m and 20 KHz is 0.0172 m respectively.

Let, the length of aluminium rod is l.

Now, firstly we calculate the speed of sound in air.

We know that,

Speed of sound (in air) =

346 =

Time taken in air = (i)

Now, speed of sound in aluminium.

We know that,

Speed of sound (in aluminium) =

6420 =

Time taken in aluminium = (ii)

Now by dividing (i) and (ii), we get:

_{= 6420/346}

=

Therefore, the ratio is 18.55:1

It is being given that,

No. of vibrations in 1 second = 100

No. of vibrations in 60 seconds = 100 × 60

= 6000

Therefore, the source of sound vibrates 6000 times in a minute.

Yes, the sound wave follows the same laws of reflection as the light does. The laws of reflection of sound are as follows:

(i) The incident sound wave, the reflected sound wave and the normal at the point of incident, all lie in the same plane.

(ii) The angle of incidence of sound wave and angle of reflection of sound wave to the normal are equal.

When sound waves reflected from a surface, the angle of incidence is equal to the angle of reflection to the normal and the incident wave, normal and reflected wave are in the same plane. This can be proved by experiment.

Thus, sound wave obeys the laws of reflection.

To hear the sound of echo depends upon the distance from source of sound and reflecting surface. The distance between both must be equal to or more than 17.2 meter. If the given distance is more than 17.2 meter then one can hear the echo sound on a hotter day also.

Although, in hotter day the velocity of sound increases, thus it is necessary to hear the sound of echo the distance should be more than 17.2 meter. If the given distance is equal to 17.2, then to hear the sound in hotter day would not be possible.

Bulb horn and Stethoscope are examples of practical applications of reflection of sound waves.

In bulb horn sound is amplified and sent to the desired direction because of reflection. In stethoscope also sound is sent to the desired direction because of its reflection characteristic.

Given,

Speed of sound, v = 339 m s^-1

Frequency, f =?

Wavelength, λ = 1.5 cm

=

= 0.015 m

We know that,

v = f × λ

339 = f × 0.015

f =

= 22600 Hz

As the frequency is 22600 Hz, that is beyond the upper limit of hearing. Hence, it will not be audible. It is actually ultrasonic sound.

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.

Given:

Distance covered by stone, s = 500 m

Initial speed, u = 0 (As the stone is initially at rest)

Time taken, t =?

Acceleration, g = 10 m/s²

We know the second equation of motion,

s =

Putting the values in the above equation we get:-

500 = 5 × t²

t²=

t² = 100

t = √100 = 10 sec

∴ the time taken by the stone to reach the pond = 10 s ..(i)

Now, we know that,

Speed of sound =

340 =

Time taken by sound to travel 500m upwards =

= 1.47 s ..(ii)

So, the required time at which splash is heard at the top = 10 s + 1.47s = 11.47 seconds

The sensation produced in the ears which enables us to distinguish between a faint sound (feeble sound) and a loud sound is called loudness of sound. Loudness depends on the amplitude of vibrations. In fact, loudness is proportional to the square of the amplitude of vibrations.

Greater is the amplitude of vibration, louder is the sound produced by it.

Bats emit ultrasonic waves from their mouth ( squeaks ). When these touch the prey, the waves get reflected back to the bat. the bat detects these waves and his brain estimates the distance and direction of the prey by the time taken, the loudness of the sound and the direction of the sound.

Objects to be cleansed are put in a cleaning solution and ultrasonic sound waves are passed through that solution. The high frequency of these ultrasonic waves detaches the dirt from the objects.

Given,

Time taken = 5 seconds

The time taken by ultrasound signal to travel from the submarine to the object will be half of this time:

=

= 2.5 seconds

We know that,

Speed =

=

= 1450 m/s

Therefore, the speed of sound is 1450 m/s.