Lecture 13: Energy and Power
1. Energy in SHM.
Consider again a mass on a spring, with its displacement given by x(t) = Asin(ωt), and look at the
energy it has.
The potential energy of a spring is 12 kx2 so for the mass it is 12 kA2 sin2 (ωt).
The kinetic energy is 12 mẋ2 = 12 mω 2 A2 cos2 (ωt) which is 12 kA2 cos2 (ωt), using the fact that ω 2 = k/m
So the total is 12 kA2 (sin2 (ωt) + cos2 (ωt)) = 12 kA2 . The energy is a constant, and is proportional to A2 ,
the square of the amplitude.
The energy changes from potential to kinetic and back again. To find the total energy one can find
either the maximum kinetic energy of the maximum potential energy - whichever is easier.
For damped oscillations for which the amplitude decays like e−γt/2 , the energy decays like e−γt .
2. Wave Energy in traveling waves
2.1 Transverse waves
Consider a wave on a string described by some f (x − vt).
A length δx has mass µδx and speed ∂f
∂t and thus kinetic energy
KE = 12 µ
∂f
∂t
2
δx
It also has potential energy. That’s a little more tricky. The section of string forms a right angled
δx, and Pythagoras’ Theorem gives the hypotenuse. So it has been
triangle with base δx and height ∂f
q ∂x
2
1 ∂f 2
stretched in length from δx to 1 + ∂f
∂x δx. This is, using the Taylor expansion, an extra 2 ∂x δx. This
stretching has been performed by a force T. So the potential energy (force times distance) is
P E = 12 T
.
So per unit length the potential energy is 21 T
∂f
∂x
∂f
∂x
2
2
δx
and the kinetic energy is 12 µ
2
±v ∂f
∂x and µv = T , so the potential and kinetic energies are equal, and the total is T
∂f
∂t
2
∂f
∂x
. But
2
∂f
∂t
is just
To find the total energy you integrate along the whole length. If the wave is Asin(kx − ωt), the kinetic
energy is 12 ω 2 µA2 cos2 (kx − ωt) and the potential energy is 12 T k 2 A2 cos2 (kx − ωt), which is the same, as
ω 2 = k 2 v 2 and v 2 = T /µ. On integration over a complete wavelength (or a complete number of wavelengths)
the cos2 gives an average of 21 , so the total energy per unit length is 12 T k 2 A2 or, equivalently, 12 µω 2 A2
So if a 20m rope of total mass 2 kg has a 100N tension applied, and someone excites waves of wavelength
1.5m and amplitude 0.2m, this has energy E = 20 × 21 × 100 × (2π/1.5)2 × 0.22 = 702J
2.2 Power
Suppose such a rope is agitated. Power has to be supplied. The energetic wave spreads down the rope
with speed v, so the power supplied must be 12 T k 2 A2 v
p
p
In the example - to know the power we also need to know the speed, which is v = T /µ = 100/0.1 =
31.6m/s. So the agitation requires (702/20) × 32 = 1090 Watts of power.
2.3 Longitudinal waves
Now consider a longitudinal wave in a spring.
2
The kinetic energy density is again KE = 12 µ ∂f
.
∂t
To find the PE, consider a length element and expand it from L to L + ∆L, integrating the extension
R ∆L
x from 0 to ∆L. The work done is 0 F dx
During the expansion the strain is given by x/L, and the force is Kx/L. So the work done is
R ∆L x
2
x2
K L dx = [K 2L
] = K ∆L
2L .
0
∆L 2
K
2
The work done per unit length is K
2 ( L ) = 2 (Strain)
Lecture 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Energy and Power
In these longitudinal waves, the strain is
∂f
∂x
so the potential energy density is P E = 12 K
∂f
∂x
2
Again, the PE and KE are equal as K = µv 2 .
For f = Asin(kx − ωt) the KE is 21 µω 2 A2 cos2 (kx − ωt) = 12 µv 2 k 2 A2 cos2 (kx − ωt) and the PE is
1
1
2 2
2
2
2 Kk A cos (kx − ωt), and v = K/µ. Again, averaging over a wavelength gives a factor of 2 and the total
1
1
2 2
2 2
energy per unit length is 2 Kk A = 2 µω A
3. Sound Waves
3.1 Displacement and pressure
If the displacement is Asin(kx − ωt) then the strain is ∂f
∂x = kAcos(kx − ωt). The force is K times the
strain, where (see previous lecture) K = γP0 A. So the pressure is γP0 kAcos(kx − ωt)
The pressure oscillates with an amplitude which depends on the displacement amplitude A and also on
the wavenumber k, or, equivalently, the wavelength/frequency/angular frequency. The amplitudes p and A
are related by
p=
2πγP0 f
A
v
So for a displacement amplitude of, say, 1 micron in air, for middle C (256 Hz) the pressure amplitude
is 2 × π × 1.4 × 105 × 10−6 × 256/344 = 0.65P a. For 512 Hz it would be twice as much, and so on.
3.2 Energy in sound waves
The kinetic energy density in the wave is 21 ρv 2 = 12 ρω 2 A2 cos2 (kx − ωt). On average the cos2 gives 12 .
The potential energy is equal (factor 2). So the total average energy density is 21 ρω 2 A2 . So this depends on
the amplitude and the frequency.
When the wave travels, in a time t over an area S, a volume Svt of wave crosses the area, so the energy
per second per area is 12 ρω 2 A2 v. This is the Power Density
So the 1 micron sound wave at middle C has a power density 12 × 1.3 × (2π × 256)2 × 10−12 × 344 =
0.58mW/m2
4. Loudness
Sound waves carry energy. The higher the power, the louder we perceive the sound. Our perception of
‘loudness’ is an example of Weber’s Law: Subjective effects are proportional to the logarithm of the stimulus.
Dealing with logarithms brings in decibels. Two sounds differ by 1 Bel if one has 10 times the power of
the other. (Named after Alexander Graham Bell, but with only one L). 2 Bels means we get 100 times as
much from one as the other. A decibel (dB) is 0.1 Bel, (a factor of 1.26). 3 dB is a factor of 2.
Sound is measured in “Perceived Noise Decibels” (though sometimes just referred to as ’decibels’). Zero
pNdB defined as 10−12 W/m2 (threshold of hearing at 1 KHz). Also as 20 µP a
10 pNdB quiet whisper
30 pNdB quiet street
60 pNdB conversation
busy traffic 70 pNdB
Thunder 110 pNdB
pain threshold 120 pNdB
So the human ear has a very impressive dynamic range of 12 orders of magnitude.
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