Propagation Of Sound Wave

Sound waves are mechanical oscillations moving through a medium such as air, water, or solid materials; they’re created by any object that vibrates and makes the surrounding medium vibrate too. The sound spreads from continuously moving one thing creating regions of high density and low compression behind itself, bringing about a waveform which we hear as noise.

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This Story also Contains

1. What Is Sound And Sound Wave?
2. Solved Examples Based on Sound Wave
3. Summary

In this article, we will discuss the sound waves and propagation of sound waves which are important when studying oscillations, and physics courses that apply to NEET and JEE Main. Most times, understanding sound waves makes it easy to understand other things like wave propagation, frequency and also wavelength. Longitudinal waves are sound waves that transport power.

What Is Sound And Sound Wave?

Sound is defined as the energy to which the human ears are sensitive is known as sound.

Sound waves always travel through any elastic material medium with a speed that depends on the properties of the medium. As sound waves travel through the air, the molecules of air vibrate to produce changes in density and pressure along the direction of motion of the wave. If the source of the sound waves vibrates as a Sine wave, the pressure variations are also like Sine waves. Because of this, the mathematical description of sinusoidal sound waves is very similar to that of sinusoidal waves on strings.

Propagation of Sound Waves

Sound is a longitudinal wave that is created by a vibrating source such as a guitar string, the human vocal cords, or the diaphragm of a loudspeaker. As a sound wave is a mechanical wave, so, sound needs a medium having properties of inertia and elasticity. To understand the propagation of sound waves. let us take an example -

Consider a tuning fork producing sound waves. When prong B moves outward towards the right it compresses the air in front of it and due to compression, the pressure in this region rises slightly. The region where pressure is increased is called a compression pulse and it travels away from the prong with the speed of sound.

Now, after producing the compression pulse, prong B reverses its motion and moves inward. This process drags away some air from the region in front of it, which causes the pressure to dip slightly below the normal pressure. This region is a decreased pressure region which is called a rarefaction pulse. Following immediately behind the compression pulse, the rarefaction pulse also travels away from the prong with the speed of sound. If the prongs vibrate in SHM then the pressure variation in the layer close to the prong also varies simply harmonically and shows SHM hence increase in pressure above normal value can be written as -

δP=δP0sin⁡ωt

Here,

δP0 is the maximum increase in value above the normal value

Now, the equation can also be written as -

δP=δP0sin⁡[ω(t−x/v)]

where, v is the wave velocity

So, the above equation shows the variation of pressure with time.

If the change in pressure is not very small then we can write the variation of pressure in the form of

Δp=Δpmaxsin⁡[ω(t−x/v)]

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Solved Examples Based on Sound Wave

Example 1: The maximum pressure variation that the human ear can tolerate is 30 N/m2. The maximum displacement for a sound wave in the air having a frequency of 103kHz is? (Use density of air =ρ=1.2kgm3 and speed of sound in air =v=343 m/s )

1) 2π3×10−2 km
2) 2×10−4πkm
3) π3×10−2 km
4) 10−43πkm

Solution:

Equation of sound wave

ΔP=ΔPmax⋅sin⁡[ω(t−xv)]
wherein
ΔP= variation in pressure at a point
ΔPmax= maximum variation in pressure
(ΔPmax )=BAK⇒A=ΔPBkv=ωkB=ρ×v2

k=ωρB⇒A=ΔPmax2πfρv=10−43πKm
(Use ρ=1.2kgm3 and v=343 m/s )

Hence, the answer is the option (4).

Example 2: The pressure wave, P=0.01sin⁡[1000t−3x]Nm−2, corresponds to the sound produced by a vibrating blade on a day when the atmospheric temperature is 0∘C.On some other day when the temperature is T. the speed of sound produced by the same blade and at the same frequency is found to be 336 ms−1. Approximate value of T (in ∘C ) is:

1) 4

2) 11

3) 12

4) 15

Solution:

Equation of sound wave

ΔP=ΔPmax⋅sin⁡[ω(t−xV)] wherein ΔP= variation in pressure at a point ΔPmax= maximum variation in pressure at 0∘CP=0.01sin⁡(1000t−3x)Nm−2V1=ωkV1=10003 at temp TV2=336 ms−1

V1V2=T1T210003336=273T⇒T=277.41k=T=4.4∘C
(Where T is in Kelvin)

Hence, the answer is option (4).

Example 3: Sound velocity is maximum in

1) H₂

2) N₂

3) He

4) O₂

Solution:

Effect of pressure on the speed of sound -

For ideal gas

Pρ=RTM= constant

wherein

With the change in pressure, the density also changes so the pressure does not affect the speed of sound.

v=γRTM⇒v∝γM

Since γM is maximum for H2, so sound velocity is maximum in H2.

Hence, the answer is the option 1.

Example 4: The ratio of densities of nitrogen and oxygen is 14:16. The temperature at which the speed of sound in nitrogen will be the same as that in oxygen at 55∘C is:

1) 35∘C
2) 48∘C
3) 65∘C
4) 14∘C

Solution:

v=γRTM⇒TNTO=MNMO (since the given velocities are the same) ⇒TN273+55=1416=78⇒TN=287 K=14∘C

Hence, the answer is the option 4.

Example 5: If the temperature of the atmosphere is increased, the following character of the sound wave is affected

1) Amplitude

2) Frequency

3) Velocity

4) Wavelength

Solution:

Effect of temperature on the speed of sound

∵Pρ=RTM∴V=γRTM
wherein
V∝T
T is in Kelvin
So velocity increases with an increase in temperature.
vαT

Hence, the answer is the option (3).

Summary

Sound waves move through air, water, or other substances by vibrating. This makes the particles in these substances move forwards and backwards with compressed and stretched regions being created along the way. Vibrations enable transmission of energy through one particle at a time thereby facilitating movement of sound waves in space and time.