Class 12 Physics Mechanical waves Notes

Class 12 Physics Mechanical waves Notes
UNIT 3
CLASS 12 PHYSICS • WAVE AND OPTICS

Mechanical Waves

Chapter 7
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Velocity of Wave

Type of Wave Velocity Symbols used in the scanned note
Longitudinal wave in liquid and gas v = √(B/ρ) B = bulk modulus of elasticity of the medium; ρ = density of liquid or gas
Longitudinal wave in solid v = √(Y/ρ) Y = Young’s modulus of elasticity; ρ = density of solid
Transverse wave on a stretched string v = √(T/μ) T = tension on string; μ = mass per unit length of string
Electromagnetic wave v = 1/√(με) ε and μ are the medium constants written in the source

i) Velocity of Longitudinal Wave in Liquid and Gas

v = √(B/ρ)
where, B = bulk modulus of elasticity of the medium
ρ = density of the liquid or gas

ii) Velocity of Longitudinal Wave in Solid

v = √(Y/ρ)
where, Y = Young’s modulus of elasticity
ρ = density of solid

iii) Velocity of Transverse Wave on a Stretched String

v = √(T/μ)
where, T = tension on the string
μ = mass per unit length of string

iv) Velocity of Electromagnetic Wave

v = 1/√(με)
where, ε and μ are medium constants as written in the source note.

Velocity of Sound in Any Medium by Dimensional Method

Sound wave is longitudinal wave. The velocity ‘v’ of sound wave in any medium depends upon the medium of elasticity ‘E’ and density of the medium ‘ρ’ through which it travels.

v ∝ Eaρb
v = kEaρb   — (i)
where, k is proportionality constant.

Dimensional equation of eqn (i):

[M0LT−1] = k[ML−1T−2]a[ML−3T0]b
[M0LT−1] = k[Ma+bL−a−3bT−2a]

Equating dimensions on both sides:

a + b = 0   — (ii)
−a − 3b = 1   — (iii)
−2a = −1   — (iv)

Solving these equations, we get:

a = 1/2,   b = −1/2

Using value of a & b in eqn (i):

v = kE1/2ρ−1/2
Here, k = 1
v = √(E/ρ)
For solid, E = Y (Young’s modulus).
For liquid and gas, E = B (Bulk modulus).

Newton’s Formula for Velocity of Sound

According to Newton, when sound wave travels through medium, the volume and pressure can be changed but temperature of the medium remains constant. Hence it obeys the condition of isothermal process. The equation of state for isothermal process is:

PV = constant   — (i)

Differentiating equation (i) on both sides:

P dV + V dP = 0
P dV = −V dP
P = −dP/(dV/V) = B
[where, B = bulk modulus of air]

Where:

v = √(B/ρ)
∴ v = √(P/ρ)

This is Newton’s formula for velocity of sound in air.

At NTP

P = 1.01×105 N/m2
ρ = 1.293 kg/m3
v = √[(1.01×105)/1.293]
v = 280 m/s

This value does not match with experimental value. Therefore Laplace corrected Newton’s formula by suggesting that during propagation of sound in air, temperature does not remain constant.

Laplace Correction

According to Laplace, when a sound wave travels through air, at the compression region temperature increases and at the rarefaction temperature decreases during the propagation of sound wave. Therefore propagation of sound wave takes place under adiabatic process.

The equation of state for adiabatic process is:

PVγ = constant

Differentiating above equation on both sides:

γPVγ−1dV + VγdP = 0
γP dV + V dP = 0
γP = −dP/(dV/V) = B

So, velocity of sound in air:

v = √(B/ρ)
v = √(γP/ρ)

This is Laplace formula.

At STP

P = 1.01×105 pascal
ρ = 1.293 kg/m3
γ = 1.4
v = √[(1.4×1.01×105)/1.293]
v = 331 m/s

This result closely agrees with the experimental value. Thus Laplace formula gives the correct value of velocity of sound in air.

Factors that Affect Velocity of Sound

  1. Effect of density of gas
  2. Effect of temperature
  3. Effect of pressure
  4. Nature of gas
  5. Effect of humidity
  6. Effect of wind

1. Effect of Density of Gas

Velocity of sound in air is given by:

v = √(γP/ρ)
where P = pressure
γ = Cp/Cv is the ratio of specific heat capacity
ρ = density of gas

From above formula for constant pressure:

v ∝ 1/√ρ

This shows that at constant pressure velocity of sound is inversely proportional to square root of density of gas.

2. Effect of Temperature

We have, v = √(γP/ρ)
Also, ρ = m/V, where m = mass and V = volume
v = √(γPV/m)

For one mole of gas equation:

PV = RT   [R = universal constant]
∴ v = √(γRT/m)

For given gas γR/m is constant:

v ∝ √T

Hence velocity of sound is directly proportional to square root of temperature.

Note:
v1/v2 = √(T1/T2)
v2/v1 = √(T2/T1)

3. Effect of Pressure

We have, v = √(γRT/m)
This means velocity of sound is independent of pressure.

4. Nature of Gas

We have, v = √(γRT/m)

For constant γRT:

v ∝ 1/√m

This shows that velocity of sound in gas is inversely proportional to square root of molecular mass of gas. Hence, velocity of sound in hydrogen gas is greater.

5. Effect of Humidity

As the increase in humidity in a gas its density decreases but we have relation:

v ∝ 1/√ρ

This shows that as decrease in density its velocity increases. Hence velocity of sound increases when humidity increases.

6. Effect of Wind

Let Vs and Vw be the velocity of sound and wind respectively and angle between them is ‘θ’ as shown in figure.

Vector diagram showing velocity of sound, wind and the angle theta between them O A Vₛ Vw θ
Effect of wind: velocity vectors of sound and wind

Now, resultant velocity of sound along OA:

V = Vs + Vw cosθ

Special Cases

Case I: θ = 0°

The velocity of wind and sound are in same direction.

V = Vs + Vw cos0°
V = Vs + Vw

In this case velocity of sound increases.

Case II: θ = 180°

The velocity of wind and sound are in opposite direction.

V = Vs + Vw cos180°
V = Vs − Vw

In this case velocity of sound decreases.

Case III: θ = 90°

The velocity of wind and sound are perpendicular.

V = Vs + Vw cos90°
V = Vs

Solved Numericals

Q.1 — Temperature for 50% Increase in Sound Velocity

At what temperature, the velocity of sound in air is increased by 50% to that at 27°C?

Let V1 be the velocity of sound in air at T1.
T1 = 27°C = 300 K
Let V1 = V
V2 = V + V of 50% = V + V/2 = 3V/2
T2 = ?

Now:

V1/V2 = √(T1/T2)
V/(3V/2) = √(300/T2)
2/3 = √(300/T2)

Squaring:

4/9 = 300/T2
T2 = 675 K

Therefore, the required temperature is 675 K.

Q.2 — Difference of Sound Speeds at −13°C and 27°C

What is the difference between the speed of longitudinal wave in air at −13°C and their speed at 27°C?

Speed at 0°C:

v = √(γP/ρ)
= √[(1.4×1.01×105)/1.293]
v = 331.2 m/s

Case I:

V1 = ?
V2 = 331.2 m/s
T1 = −13°C = 260 K
T2 = 0°C = 273 K
V1/V2 = √(T1/T2)
V1/331.2 = √(260/273)
V1 = 323.2 m/s

Case II:

V′1 = 331.2 m/s
V′2 = ?
T′1 = 0°C = 273 K
T′2 = 27°C = 300 K
V′1/V′2 = √(T′1/T′2)
331.2/V′2 = √(273/300)
V′2 = 347.19 m/s

Therefore, difference:

V′2 − V1 = 347.19 − 323.2
= 23.99 m/s

Q.3 — Distance of Lightning Strike

In a stormy day, a boy observes a lightning flash which is followed by a thunder 3 sec later. How would you estimate the distance of the lightning strike from the boy? (Given velocity of sound on that day = 332 m/s, velocity of light = 3×108 m/s.)

Velocity of light, Vl = 3×108 m/s
Velocity of sound, Vs = 332 m/s
Time interval = 3 sec

Now:

ts − tl = 3 sec
d/Vs − d/Vl = 3
d(1/332 − 1/(3×108)) = 3
(3×108 − 332)/(332×3×108) = 3/d
d = 997 m

Q.4 — Velocity from Frequency and Wavelength

A source of sound of frequency 550 Hz emits waves of wavelength 660 mm in air at 20°C. What is the velocity of sound in air at this temperature? What would be the velocity of sound at 0°C?

Given, T1 = 20°C = 293 K
f = 550 Hz
Wavelength, λ = 660 mm = 0.66 m
Initial velocity, V1 = λf = 0.66×550
V1 = 363 m/s
Final temperature, T2 = 0°C = 273 K
V2 = ?
V2/V1 = √(T2/T1)
V2 = V1√(T2/T1)
= 363√(273/293)
V2 = 350.39 m/s

Q.5 — Temperature When Sound Velocity Doubles

Find the temperature at which the velocity of sound in air is double the velocity of sound in air at 0°C. (Velocity of sound in air at 0°C = 330 m/s.)

Given velocity of sound in air at 0°C = V0 = 330 m/s
Velocity at temperature T, V = 2V0

Now:

V0/V = √(T0/T)
V02/V2 = T0/T
T = 273 × V2/V02
= 273 × (2V0)2/V02
= 273 × 4
T = 1092 K

So, required temperature shown by subtracting 273 in the source:

1092 − 273 = 819
Required temperature = 819°C
The handwritten final unit after “819” is not fully clear, but the calculation shown is 1092 − 273 = 819.

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