Mechanical waves

                                   Answer Questions :- 

1.  Why sound made at a distance can be heard distinctly at night than in the day time?
>At night, the atmospheric air becomes moist than at day time. So, density of air at night is less
than at day time. But the velocity of sound is inversely proportional to the square root of density i.e.,

 V  ∞ 1/ . So, velocity of sound at night is more than at day time due to reduced density of air. Hence, sound made at a distance can be heard distinctly at night than in daytime.


2. Is velocity of sound more in damp air or in dry air? Explain.
> In the wet day, the atmospheric air becomes moist than on a dry day. Pressure of moist air
reduces its density. We know that the velocity of sound in air is inversely proportional to the
square root of the density i.e. V ∞ 1/ . Therefore, decrease in density causes the velocity of sound to increase. Hence, sound heard better on wet day than on a dry day.

3.  Do sound waves need a medium to travel from one point to other point in space? What properties of the medium are relevant?
> Yes, sound waves are mechanical waves which need a medium to travel from one point to other point in space. In this case, the disturbance is handed over from one particle to another particle of the medium Elastic property should be possessed by the medium. For the particle to gain kinetic energy, medium should possess inertia. The velocity of sound in the elastic medium is given by V= √(E/ρ)
where E is the elasticity of medium and p is its density. Thus, elasticity, pressure and inertia are the properties on which sound waves depend.


4.  Although the density of solid is high, the velocity of sound is great in solid, explain.
> Sound wave is a longitudinal wave. The velocity of a longitudinal wave (in general) in a
medium is given by the relation by V= √(E/ρ) where E is the modulus of elasticity and ρ is the density of the medium. From this relation, it is clear that v not only depends on p but also on
E. So, the velocity of propagation of any longitudinal wave in any medium depends on the
ratio E/ ρ. For a solid, the ratio is much greater than that for a gas or a liquid. Hence, the velocity of longitudinal wave (eg sound wave) in solids is greater than in liquids and gases.


5.  Velocity of sound increases on a cloudy day. Why?
> In a cloudy day, the air become moist and humidity increases which results to decrease the
density of air. Since, velocity of sound is inversely proportional to square root of density of
medium (i.e. v∝1/ ), the velocity of sound increases on a cloudy day.

 

6. Sound travels faster in metal than in air, why?
> Sound wave is a longitudinal wave. The velocity of a longitudinal wave (in general) in a medium is given by the relation V= √(E/ρ)where E is the modulus of elasticity and is the density of the medium. The modulus of elasticity for metal is Young's modulus of elasticity (Y) , then velocity of sound in metal is, V= V= √(y/ρ)The value of  Y/density for metal is to much higher than B/density for air. So, the velocity of sound wave in metal is greater than in air and hence sound travels faster in metal than in air.

7. When sound waves travel through a medium, does the temperature at various
points remain constant? Explain.
> No, when the sound waves travel in a medium, the temperature at various points does not remain same. According to Laplace during the propagation of sound wave the formation of compression and rarefraction is so rapid that the heat exchange is zero and the temperature at different points is different and hence the process is adiabatic process.

8. Discuss the effect of pressure, temperature and density of elastic medium on the velocity of sound.
> The velocity of sound in an elastic medium is given by V= √(E/ρ)where E is elasticity and p is its density.
For air medium, E=B = YP, where y is constant called the ratio of molar heat capacities of gas at constant pressure to that at constant volume and P is pressure
So, V= √(yP/ρ)
1. Effect of pressure: The velocity of sound in air is
V= √(yP/ρ)
For one mole of gas, equation of state is
PV = RT
At a constant temperature,
PV = constant
Since, V = M/p
therefore, p ( M/p) = constant P/ρ = constant
Hence, V= √(yP/ρ)= constant
Thus, the velocity of sound in a gas is independent of the pressure of the gas provided temperature remains constant.

ii. Effect of temperature: For one mole of a gas, the equation of state is
PV =  RT
If M and ρ be the molecular weight and density of the gas, then,
V = M/ρ

PM/ρ=RT=>p/ρ=RT/M
For given gas. Y, Rand M are constants.
V= √T
Hence, the velocity of sound in a gas is directly proportional to the square root of its absolute temperature.

iii. Effect of density: Consider two gases having same value of y at the same pressure P but having densities p1 and p2 respectively. Then, the velocity of sound in them is
V1= √(yP/)

V2= √(yP/)                                       (Here,we consider one mole of each gas such that PV=RT)

V1/V2 = √( ρ2/ ρ1)=V∝ √ρ


Hence, velocity of sound in a gas is inversely proportional to the square root of the density of the gs.

iv. Effect of mass: Let us consider M1 and M2 be molecular weight of the gases, then at constant temperature, the velocity of sound in them is
V1= √(yRT/M1)

V2= √(yRT/M2)                        [Here, for a given gas at constant temperature, R. Tandy are  constants)
V1/V2 = √( M2/ M1)=V∝ 1/√M
Hence, velocity of sound in a gas is inversely proportional to the square root of the molecular mass of the gas.

 

9.  Derive an expression for the velocity of sound in a medium by dimensional method. Discuss the effect of change in pressure and temperature on the velocity of sound in air.
> The velocity of sound wave in an elastic medium depends upon the elasticity and density of medium.
i.e. v is directly proportional to Ex py where E is elasticity of medium and p is density of medium
Now,
V = KEx Py -------(i)
where k is a dimensionless proportionality constant
Here,
Dimension of elasticity, E = [ML-1 T-2)
Dimension of velocity, V = [LT-1)
Dimension of density, p = [ML-3)
From dimensional analysis, from equation (i), we get
[LT-1) = (ML-1 T^-2)^x . [ ML-3]^Y
or [LT^-1) = ( Mx+y L-x-3y T-2x ]---------(ii)
Equating the powers of like terms, we get
0 = x+y..(iii)
1 = -x - 3y--(iv)
-1 -=  2x-------(v)
Solving these equations, we get
X = ½,   Y= -1/2

Therefore, V = KE1/2 p-1/2 = k=√(E/ρ)  

                                    
From mathematical analysis, k = 1, then velocity of sound in medium is V= √(E/ρ)  



Numerical Problems
1. The interval between the flash of lighting and the sound of thunder is 2  seconds. when temperature is 10°C. How tar is the storm in the velocity of sound in air at 0c is 330 ms^-1?
> Solution,
Given, the time period (t) = 2 sec

Let,
   First case:
Velocity (v1) = 330 ms-1
Temperature (T1) = 0°C = 2713 K
Second case :
Velocity (v2) = ?
Temperature (T2) = 10°C = 283K
Since, V1/V2= √(T1/T2)  

Or, 330/v­2= √(273/283)  
Or, 330/v2 = 0.98
V2 = 335.98 ms-1
Again, Using,Distance =  velocity X  time = 335.98 x 2 = 671.96 = 672m
Thus, the storm is 672 m far.


 

2. A man standing at one end of a closed corridor 57 m long blow a short blast on a whistle. He found that the time from the blast to the sixth echo was 2 seconds. If the temperature was 17°C, what was the velocity of sound at 0°C?
> Solution
Here,
The distance travelled by sound between the blast and the first echo =  2X57m
Time (t) = 2 sec
Temperature (Θ) = 17°C= 290 K
Total distance travelled in 2 sec = 6×2 ×57 m
Velocity of sound at 17°C
VΘ =6 x 2 x 57 /2 = 342 m/s
Now, let the velocity of sound at 0c be V0

Since,Vθ/V0= √(273+θ/273)

= √(273+17/273)

=V­0=√(273/290)×Vθ= √(273/290)×342=331.8        

The velocity of sound at OC =  331.8 m/sec.

 

3. At what temperature, the velocity of sound in air is increased by 50% to that at 27C?
> Solution
 Let, v be the velocity of sound at 27°C Then, at TC, the velocity of sound be
V2 = ( v + 50v/100)
Here,
T1 = 27c = 300k     V1 = v

T2 = (T + 273)K and ,  V2 = v+ 50v/100 = 3v/2
Now, we have,
V1/V2= √(T1/T2)  
V/3V//2= √(300/P+273)  

Or,2/3= √(300/P+273)  

Or,4/9=(300/P+273)
or, T+ 273 =  675
 T = 402°C
Thus, the required temperature is 402°C

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