Lo #4

SOUND WAVES (15-4)

Victoria Purcell Student #: 26656141

• When sound passes through a medium, there are alternating areas of low pressure and high pressure due

to the oscillations of molecules• High pressure (Compression) : medium is compressed• Low Pressure (Rarefraction) : medium is stretched

• The molecules are displaced parallel/antiparallel to the direction of propagation

SOUND WAVES ARE LONGITUDAL WAVES

• The speed depends on the characteristics of the medium

As you recall from Chapter 14, the speed of waves on springs is determined by the mass density (μ) and tension T

• v =√T/μ

For air, the mass density is simply the density of the medium ρ

Another flashback to Chapter 14, the spring constant k defines how much the length changes with force. Bulk modulus B is similar to k as it is

defined by how much the volume changes when the pressure exerted is changed.

B = -(Δp/ΔV)*V

There is a negative sign in this formula because the change in volume is opposite in sign to the change in pressure

We can take this information and plug it into the equation for the velocity of a wave on a string and we get the speed of sound as v=√B/ρ

SPEED OF SOUND

If the wave is travelling in the positive x direction then

s (x,t)=smcos(kx-ωt+φ)

with sm = maximum displacement from equilibrium,

wave number k= 2π/λ,

Angular frequency ω= 2πf

And the phase constant φ

And

• The relationship between velocity, frequency and wavelength• v=fλ=ω/k

DISPLACEMENT AMPLITUDE

• Each medium has an ambient pressure, and each increase or decrease relates to a pressure variation

• Displacement variations and pressure variations have the same periodicity (both in time and space)

• A single frequency sound wave varies sinusoidally

In regions of high pressure, maximum pressure is when the displacement is equal to zero

Similarly, in regions of low pressure, minimum pressure is when the displacement is equal to zero

PRESSURE VARIATIONS

Change in displacement: Δs = (ϑs/ϑx)*Δx

This is a partial derivative because s depends on both x and t

Change in volume: ΔV=AΔs

Which, from the previous equation, means ΔV=A(ϑs/ϑx)Δx

In the equation we multiply the change in displacement by the cross-sectional area to find the change in volume and

rearrange it to get

(ΔV/AΔx)= ΔV/V = (ϑs/ϑx)

In order to relate the volume variations to pressure variations, we substitute volume into the bulk modulus

formula to achieve

(ΔV/V) = -Δp/B = (ϑs/ϑx) which is algebraically rearranged to

Δp=-B(ϑs/ϑx)

FORMULA DERIVATION

By taking the partial derivative of the displacement formula we can achieve

(ϑs/ϑx)= -ksmsin(kx-ωt+ϕ)

And since -B(ϑs/ϑx)=Δp,

Δp=Bksmsin(kx-ωt+ϕ)

CONTINUED

Δp=Bksmsin(kx-ωt+ϕ

The amplitude is Δp=Bksm

AMPLITUDE

The difference between the wave and pressure formulas is that the wave uses a cosine function and the pressure

formula uses a sinusoidal function.

Each function has the same wavelength, period and wave speed except they are π/2 out of phase from each other

WAVE AND PRESSURE FORMULAS

Intensity is the power delivered per unit area

I = P/A

P is the rate at which the wave delivers energy and A is the area the wave is influencing.

The power of a mechanical wave is described as

Pavg=(1/2)μvω2A2

Sound wave density is represented by ρ and the amplitude is sm, giving us the equation for the intensity of a sound

wave as

I=(1/2)ρvω2sm2

INTENSITY

1.) Compared to sound of a 44 decibels, a sound of 34 decibels is

a) 5 times less intense

b) 10 times less intense

c) 10 times more intense

d) 100 times less intense

2.) True or False: The bulk modulus is related to the change in pressure but not the change in volume.

PRACTICE

3.) If this is the displacement graph of a sound wave,

then what is the graph of the pressure change for this wave

PRACTICE

4.) True or False: A in the formula for intensity represents the cross-sectional area of the wave

5.) Show how the velocity = fλand ω/k

6.) What is the velocity of a wave with a 5 mm wavelength that oscillates three times completely in one second?

7.) If the change in a volume of 20 cm2 is 12 cm2 and the change in x is 5 cm, what is the change in displacement?

PRACTICE

1.) b: 10 times less intense

2.) False: Δp=-B(ϑs/ϑx) and ΔV/V = (ϑs/ϑx) relating bulk modulus to both the change in pressure and change in

volume

3.) The funcion of the pressure is

π/2 radians out of phase with the displacement function but have the same wavelength and period

ANSWERS

4.) False: A is not the cross-sectional area of the wave itself, but the area that the wave is affecting.

5.) ω= 2πf and k=2π/λ

If v= ω/k then v = (2πf)/(2π/λ)

v = (2πf)*(λ/2π)

The 2π cancel out and we are left with

v = fλ

6.) If the wave oscillates 3 times in a second, the frequency is 3 Hz.

v= fλ and the wavelength was stated to be 5 mm so the velocity is

v= 3Hz * (5mm/1000mm)= 3Hz*(0.005m) = 1.5e-2 m/s

ANSWERS

7.) ΔV/V= (ϑs/ϑx) and

(ϑs/ϑx)*Δx = Δs

So ΔV/V= 12 cm2/20cm2 =(ϑs/ϑx)

So (ϑs/ϑx) = 0.6

And with the change in x being 5 cm,

0.6*5cm= 3 so Δs=3

ANSWERS

Physics for Scientists and Engineers “An Interactive Approach”. J. Iqbal, F. Mansour, M. Milner-Bolotin, P. Williams and R. Hawkes. Nelson Education Ltd. 2014

SOURCES