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Re: Standing waves.

Quoting Ludwik Kowalski <kowalskil@MAIL.MONTCLAIR.EDU>:

How to explain this in an introductory physics course?

Standing waves apparatus from Pasco is
used to observe loops established on a
string at frequencies corresponding to the
known L, tension, and mu. In preparing for
tomorrow's lab I saw that the amplitudes of
loops decrease monotonically (from about
4 cm when the lambda is 2*L to about 0.2
cm when lambda is 9 times shorter.)

Good observation.

Why is it so?

Good question; see analysis below.

My setup: Pasco vibrator SF-9324
Soft threaded string mu=0.00160 kg/m and
L=126.5 cm,
Pasco generator (amplitude was constant (in
the middle of the range) only frequencies were
changed from about 14 Hz to 128 to observe
1, 2, 3, 4, ...9 loops.

Good detailed report!

One more detail: I assume "observations" were taken on-resonance.

Why do the amplitudes of standing waves
decrease while the signal amplitude of the
source remains the same?

For identical amplitudes the string would
have to be longer with 9 loops than with
one loop.

That's a plausible hypothesis, but it doesn't fit the observations.
The law of conservation-of-string will produce some nonlinearity
(resonant frequency depending on amplitude) but doesn't explain
the basic fact, to wit:

I think any analysis must deal with dissipation. In the absence
of dissipation, on-resonance, the wave equation predicts
*infinite* amplitude.

There is a natural intuition, a sort of "scaling law", suggesting
that the amplitude of the wave should scale like the amplitude of
the drive. This is not entirely wrong, but significantly incomplete.
Scaling laws won't tell you about dimensionless factors ... such as

One way to get started on a proper analysis is to consider the
energy budget.

-- On one side of the balance sheet, energy is being lost at a
rate that depends on
amplitude *and* frequency. Probably aerodynamic drag is
dominant, in which case the drag force should scale roughly like
the square of the local velocity, and the power should scale
like the cube.

-- On the other side of the balance sheet, to understand the
way energy is being put into the string you must realize that
in the presence of dissipation you haven't got the ideal
standing wave. Some fraction of the energy is, in effect,
being put into a running wave that runs off and never returns.

As usual, for any particular frequency the wave equation
reduces to a plain old harmonic oscillator. Use the wave
equation to find the HO equation, and then carry out the
rest of the analysis using the latter. You will find a
frequency-dependent Q.

As with any other driven damped harmonic oscillator, you will
find that exactly at resonance, the response will be a factor
of Q bigger than the drive (and 90 degrees out of phase).