Different Expressions for Wave Speed
- Wave speed V= ω/k. By definition of ω= (2 π)/T where T is period of the wave.
- Also k=(2 π)/λ .Therefore v=(2 π)/(T)/(2 π)/(λ) =λ/T
- v= λv
- As wavelength of a wave increases as a result frequency of the wave decreases as a result speed of wave is constant.
- Wave speed is determined by the properties of the medium.
Dimensional Analysis to show how the speed is related to mass per unit length and Tension
- μ = [M]/[L] = [ML-1] (i)
- T=F=ma =[M][LT-2] = [MLT-2] (ii)
- Dividing equation (i) by (ii) :- [ML-1]/[MLT-2] = L-2T-2 =[T/L]2 =[TL-1]2 =1/v2
- Therefore μ/T = 1/v2
- v=C√T/ μ where C=dimensionless constant
- Conclusion: v depends on properties of the medium and not on frequency of the wave.
Problem:- A steel wire 0.72 m long hasa mass of 5.0 ×10–3 kg. If the wire is under a tension of 60 N, what is the speed oftransverse waves on the wire?
Answer: -Mass per unit length of the wire,0.72 m
μ = 5.0×10−3 kg
= 6.9 ×10–3 kg m–1
Tension, T = 60 N
The speed of wave on the wire is given by,v=C√T/ μ
√ (60)/ (6.9×10-3kgm-1) = 93ms-1
Problem:- A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20.0 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end?
Mass of the string, M = 2.50 kg
Tension in the string, T = 200 N
Length of the string, l = 20.0 m
Mass per unit length,μ=M/l=2.50/20=0.125kgm-1
The velocity (v) of the transverse wave in the string is given by the relation: v=√T/μ
Therefore, time taken by the disturbance to reach the other end, t =l/v=20/40=0.50s
Problem:- A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the tension in the wire so that speed of a transverse wave on the wire equals the speed of sound in dry air at 20 °C = 343 m s–1.
Length of the steel wire, l = 12 m
Mass of the steel wire, m = 2.10 kg
Velocity of the transverse wave, v = 343 m/s
Mass per unit length,μ=m/l= 2.10/12 =0.175kgm-1
For tension T, velocity of the transverse wave can be obtained using the relation:
= (343)2 × 0.175 = 20588.575 ≈ 2.06 × 104 N
Speed of a longitudinal wave in a stretched string
Longitudinal wave speed determined by:
- Density – Longitudinal wave is formed due to compressions (particles very close to each other) and rarefactions (particles are far from each other).
- At certain places it is very dense and at certain places it is very less dense.So density plays a very important role.
- It is denoted by ρ.
- Bulk modulus– Bulk modulus tells how does the volume of a medium changes when the pressure on it changes.
- If we change the pressure of compressions orrarefactions then the volume of the medium changes.
- It is denoted by B.
Dimensional Analysis to show how the speed is related to density and bulk modulus
- ρ =mass/volume= [ML-3]
- B = – (Change in pressure (ΔP))/Change in volume(ΔV/V))
- ΔV/V is a dimensionless quantity as they are 2 similar quantities.
- ΔP =F/A = ma/A =[M][LT-2]/[L2] = [ML-1T-1]
- Dividing ρ/B= [ML-3]/[ML-1T-2] = [L-2T2]=[TL-1]2 =1/v2
- Therefore ρ/B = 1/v2 or v2=B/ ρ
- v= C √ (B/ ρ) where C=dimensionless constant
- In case of fluids :- v= C √ (B/ ρ)
- In case of solids :- v= C √ (Y/ ρ)
Problem:- Estimate the speed ofsound in air at standard temperature andpressure. The mass of 1 mole of air is29.0 ×10–3 kg.
We know that 1 mole of any gasoccupies 22.4 litres at STP. Therefore, density
of air at STP is:ρo = (mass of one mole of air)/ (volume of onemole of air at STP)
29.0 x 103 kg/22.4×103 m3
= 1.29 kg m–3
According to Newton’s formula for the speedof sound in a medium, we get for the speed of
sound in air at STP,
v = [(1.01×105Nm-2)/ (1.29kgm-3)]1/2