OpenStax-CNX module: m12793
1
Energy in a mechanical wave
∗
Paul Padley
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 2.0
Abstract
Calculate the energy in a wave on a string.
1 Energy Transport
Figure 1
∗ Version
1.2: Aug 19, 2008 12:04 pm -0500
† http://creativecommons.org/licenses/by/2.0/
http://cnx.org/content/m12793/1.2/
OpenStax-CNX module: m12793
2
Lets calculate the energy in a wave of a string:
Consider a fragment of string so small it can be considered straight, as is shown in the gure
The the kinetic energy is K = 21 mv 2 for the string fragment m = µdx
note:
Why is this dx and not ds? Lets consider that the string is not perturbed, then it is
horizontal and has mass as given. When the string is perturbed it stretches a little bit - but the
mass does not increase.
So we have
2
∂y
1
K = µdx
2
∂t
and using this we can dene the energy per unit length, ie. the kinetic energy density:
2
dK
1
∂y
= µ
dx
2
∂t
When the string segment is stretched from the length dx to the length ds an amount of work = T (ds − dx)
is done. This is equal to the potential energy stored in the stretched string segment. So the potential energy
in this case is:
U = T (ds − dx)
Now
1/2
ds = dx2 + dy 2
2 1/2
∂y
= dx 1 + ∂x
Recall the binomial expansion
n
(1 + A) = 1 + nA +
so
n (n − 1) (n − 2) A3
n (n − 1) A2
+
+ ...
2!
3!
2
1 ∂y
dx
2 ∂x
2
∂y
1
dx
U = T (ds − dx) ≈ T
2
∂x
ds ≈ dx +
or the potential energy density
2
dU
1
∂y
= T
dx
2
∂x
To get the kinetic energy in a wavelength, lets start with
y = Asin
2πx
− ωt
λ
∂y
= −ωAcos
∂t
Lets evaluate it at time 0.
so
http://cnx.org/content/m12793/1.2/
2πx
− ωt
λ
2πx
λ
dK
1
2πx
= µω 2 A2 cos2
dx
2
λ
∂y
|
= −ωAcos
∂t t=0
OpenStax-CNX module: m12793
3
now integrate
Rλ
K = 0 dK
dx dx
R
1
2 2 λ
= 2 µω A 0 cos2 2πx
dx
λ
In order to do this integral we use the following trig identity:
cos2 A =
cos2A + 1
2
so we get
K = 12 µω 2 A2
x
2
+
λ
4πx
8π sin λ
2 2
= 14 µλω A
In similar fashion the potential energy can be found to be
U=
1
µλω 2 A2 .
4
Deriving this will be assigned as a homework problem
So
E =K +U =
Power
2 2
1
∆E
2 µλω A
∆t =
τ
= 12 µω 2 A2 v
P =
Where I have used τ = 1/ν and λν =v thus τ = λ/v
http://cnx.org/content/m12793/1.2/
1
µλω 2 A2
2
|λx=0