A powerpoint for explaining damped oscillations and the equations associated with it. Includes a sample question in the powerpoint. For physics 101 learning object 2.
2. Damped Oscillations
™ Typically, when something is oscillating, there is an
opposing force (friction or drag) acting on the
oscillation and causing it to slow down and come to a
stop.
™ If you were to swing a ball attached to the end of a
string hanging on the ceiling, you would eventually see
that it comes to a stop. This is called damping!
3. Equations
™ Displacement as a function of time:
x(t) = Ae-bt/2mcos(ωDt+φ) where
™ The first half of the function for
displacement is an equation describing how the
amplitude changes (shrinks as a result of damping):
A(t) = Ae-bt/2m
™ The cosine half of the equation describes the
oscillation.
b = drag coefficient with units of kg/s
So, the bigger b is, the bigger the drag force is.
4. Frequency
™ Mass on a spring:
™ Simple pendulum:
™ Drag affects frequency, so it’s different in damped
oscillations:
Natural Frequencies
5. Frequency of Damped
Oscillations
™ This is the equation for frequency in damped
oscillations:
™ Where ωo changes depending on whether you’re
working with a simple pendulum or a mass on a spring.
You use one of those natural frequencies.
6. Question
™ Given a system where a mass of 120 g is oscillating on
a spring, the spring constant k 97 N/m and the drag
coefficient b is 0.18 kg/s, how long will it take for the
amplitude to decrease to 25% of its original amplitude?
120 g
b=0.18 kg/s k=97 N/m
7. Solution
™ This is the equation for the displacement of a damped
oscillator:
x(t) = Ae-bt/2mcos(ωDt+φ)
™ But since the question is only asking about the
amplitude, we only need to use this part of it:
A(t) = Ae-bt/2m
8. Solution (cont’d)
™ The question asks when the amplitude will be 25% of
the original, so we go from this: A(t) = Ae-bt/2m
to 0.25Ao= Aoe-bt/2m
™ Here you can see that you don’t actually need to know
the initial amplitude because there’s a term on both
sides, so they cancel out and you get this:
0.25=e-bt/2m
™ Since we’re solving for time, we need to bring the
power down. We do so by logging both sides.
9. Solution (cont’d)
™ Logging both sides gives you this:
ln0.25 = (-bt/2m)(lne)
™ The natural log of e is just 1 so you get:
ln0.25 = -bt/2m
™ Now you can isolate t and solve for time:
™ ln0.25 gives you a negative number, which cancels with
the negative in front of the negative 2 to give you a
positive number overall. Kg cancel out to give you time
in seconds. Notice that k wasn’t used at all.
Don’t forget to
change from g
to kg!
