Re: Simple question about wave energy

Mike Christie wrote:
Sam Wormley wrote:
Why should the frequency affect the energy of mechanical waves?

It seems logical to me, because the gadget that wiggles the end of the
string from side to side has to move twice as fast to double the
frequency, therefore (presumably) it consumes more power. Where is that
additional energy going, if not into the wave itself? (Disregarding
friction, etc.)

I've fiddled with those chains they have in banks to tie pens down to
the counter. You can create transverse waves in those chains; you have
to work harder to create higher frequency waves. Doesn't "work harder"
mean "more energy is going into the wave"?

Not necessarily. Remember that if there were no friction, the chain
would go on waving forever. That tells you that once you've got the
waving started, most of the power you're using is going into
overcoming friction.

Try: http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

http://www.ndt-ed.org/EducationResources/CommunityCollege/Ultrasonics/cc_ut_index.htm


Energy in a mechanical wave
http://cnx.org/content/m12793/latest/

My math is rusty, but if I understand this last page correctly it is
deriving the kinetic energy in a single wavelength. If that's right,
that would explain my confusion -- the energy in a single wavelength
depends only on amplitude, but the energy in a string will depend on
frequency too, since a higher frequency waves will have more wavelengths
in a given length of string. Is that correct?

You're right that it is the energy in a single wavelength (E in their
notation) they are finding, but note that their result is:

E = (1/2) * (mass per unit length) * (wavelength) * (frequency /
(2*pi))^2 * (amplitude)^2

You can factor out a factor of (wavelength * frequency) and apply
velocity = wavelength * frequency:

E = (1/2) * (mass per unit length) * (velocity) * (frequency) *
(amplitude)^2 / (2*pi)^2

So the energy in one wavelength does depend on the frequency.

To get the energy in the whole string, we multiply the first equation
by the number of wavelengths on the string, which would be (string
length / wavelength), getting:

(1/2) * (mass per unit length) * (string length) * (frequency /
(2*pi))^2 * (amplitude)^2

So both depend on frequency.

--
Jim E. Black

.

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