Sound Waves Concept Page - 4

Law

Variation of speed of sound with temperature

Velocity (v) = γp/ρ−−−−√$\sqrt{\gamma p/\rho }$

as pρ=RTM${\displaystyle \frac{p}{\rho }}={\displaystyle \frac{RT}{M}}$
Therefore, Velocity (v) = γRT/M−−−−−−−√$\sqrt{\gamma RT/M}$

This expression shows variation of velocity of sound with temperature.

Formula

Variation of speed of sound with pressure

Variation in speed of sound with pressure is given by Laplace's equation:

Velocity (v) = γp/ρ−−−−√$\sqrt{\gamma p/\rho }$
Here p$p$ is pressure.
As it's clear from this equation that how velocity of the sound is in relationship with pressure.

Example

Use the variation of speed of sound with density

Velocity of sound is:
v = γp/ρ−−−−√$\sqrt{\gamma p/\rho }$

Example: The velocity of sound is vs$v_s$ in air . If the density of air is increased to 4 times , then find the new velocity of sound ?

Solution:
vS=γairpρ1−−−−−√$v_S=\sqrt{{\displaystyle \frac{{\gamma }_{air}p}{{\rho }_1}}}$
v′S=γairpρ2−−−−−√$v_S^{\prime }=\sqrt{{\displaystyle \frac{{\gamma }_{air}p}{{\rho }_2}}}$
v′SvS=ρ1ρ2−−−√=ρ14×ρ1−−−−−−√=12${\displaystyle \frac{v_S^{\prime }}{v_S}}=\sqrt{{\displaystyle \frac{{\rho }_1}{{\rho }_2}}}=\sqrt{{\displaystyle \frac{{\rho }_1}{4\times {\rho }_1}}}={\displaystyle \frac{1}{2}}$
v′S=vS2$v_S^{\prime }={\displaystyle \frac{v_S}{2}}$

Definition

Variation of speed of sound with humidity

Humidity has a small but measurable effect on the speed of sound (causing it to increase by about 0.1%0.6%), because oxygen and nitrogen molecules of the air are replaced by lighter molecules of water.For example, sound travels about 0.35 percent faster in 100 percent humidity (very humid air) than it does in 0 percent humidity (completely dry air).

Definition

Calculate the effect of velocity of medium on speed of sound

The speed of sound is not always the same. Remember that sound is a vibration of kinetic energy passed from molecule to molecule. The closer the molecules are to each other and the tighter their bonds, the less time it takes for them to pass the sound to each other and the faster sound can travel. It is easier for sound waves to go through solids than through liquids because the molecules are closer together and more tightly bonded in solids. Similarly, it is harder for sound to pass through gases than through liquids, because gaseous molecules are farther apart. The speed of sound is faster in solid materials and slower in liquids or gases. The velocity of a sound wave is affected by two properties of matter:
1. The elastic properties
2. Density.
The relationship is described by the following equation.
Cijρ−−−√$\sqrt{{\displaystyle \frac{C_{ij}}{\rho }}}$
where Cij$C_{ij}$ = The elastic properties and ρ$\rho$= Density 

Example

For a given wave function infer the different parameters of a wave

Example: The transverse displacement of a string (clamped at its both ends) is given by y(x,t)=0.06sin(2πx3)cos(120πt)$y(x,t)=0.06{\displaystyle \sin \left(\frac{2\pi x}{3}\right)\cos \left(120\pi t\right)}$, where x and y are in m and t in s. The length of the string is 1.5 m and its mass is 3.0×10−2$3.0\times {10}^{-2}$ kg. 
Answer following :
(a) Does function represent a traveling wave or a stationary wave?
(b) Interpret the wave as a superposition of two waves traveling in opposite directions. What is the wavelength, frequency, and speed of each wave?
(c) Determine the tension in the string.

Solution:
(a)Traveling wave is given by y(x,t)=Asin(ωt±kx+ϕ)$y(x,t)=A\sin (\omega t\pm kx+\varphi )$
Standing wave is given by y(x,t)=Asin(kx)cos(ωt)$y(x,t)=A\sin (kx)\cos (\omega t)$Hence, this is an example of standing wave.

(b) y(x,t)=0.06sin(2πx3)cos(120πt)$y(x,t)=0.06\sin \left({\displaystyle \frac{2\pi x}{3}}\right)\cos (120\pi t)$
y(x,t)=0.03sin(2πx3+120πt)+0.03sin(2πx3−120πt)$y(x,t)=0.03\sin ({\displaystyle \frac{2\pi x}{3}}+120\pi t)+0.03\sin ({\displaystyle \frac{2\pi x}{3}}-120\pi t)$
Wavelength, λ=2πk=2π2π/3=3 m$\lambda ={\displaystyle \frac{2\pi }{k}}={\displaystyle \frac{2\pi }{2\pi /3}}=3m$
Frequency, f=ω2π=120π2π=60 Hz$f={\displaystyle \frac{\omega }{2\pi }}={\displaystyle \frac{120\pi }{2\pi }}=60Hz$
Speed, v=fλ=180 m/s$v=f\lambda =180m/s$

(c)Speed of a wave in a string is given by: v=T/μ−−−−√$v=\sqrt{T/\mu }$
⟹T=v2m/l$⟹T=v^{2}m/l$
T=1802×3×10−21.5=648 N$T={\displaystyle \frac{{180}^2\times 3\times {10}^{-2}}{1.5}}=648N$

Definition

Comparison between speed of sound and speed of light

Speed of sound and speed of light there is a huge difference. 
speed of Light = 3×108$3\times {10}^8$ms${\displaystyle \frac{m}{s}}$
speed of Sound = 340 ms${\displaystyle \frac{m}{s}}$
No object can travel Faster then sound that is why in hollywood movies object go to lighting fast speed.

Example

Consequence of huge difference in speed of light and sound

Consequences of huge difference in speed of sound and light are:

• Thunder is heard much later than the lightning.
• Spectator of a race feels that the racers begin before the gunshot.

Definition

Speed of sound waves and electromagnetic waves

Speed of sound waves is different from speed of electromagnetic waves. Speed of both kinds of waves changes with change in medium. However speed of electromagnetic waves (maximum 3×108 m/s$3\times {10}^{8}m/s$ in air) is much more than the speed of sound waves (maximum 330 m/s$330m/s$ in air), almost around a million times as fast.
Note: Light is an electromagnetic wave.

Definition

Characteristics of a sound wave

Characteristics of sound waves are:

• It is produced by the periodic disturbance at a point in the medium.
• The particles vibrate about their mean positions and do not actually travel with the wave.
• Energy is transferred with a constant speed from one place of medium to the other place.
• It travels in the form of compression and rarefaction.

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