Chapter 30 1

PHY 184 HW: Damped Oscillations homework problem

Chapter 30 2

Chapter 30 3

Read quantities of the graph !  The curve gives you two important pieces of information: The period and the amplitude of the sine wave. •  Use the period to determine the angular frequency ω. •  Use the decay of the amplitude to determine the exponential and

from that find the resistance R=RL. !  The sine wave peaks at t0 = 0 ms and at t1 = 0.7 ms

•  ms = milliseconds, 10-3 s •  This corresponds to 8 periods •  Period T = 0.7 ms / 8 = 0.0000875 s •  Thus the angular frequency is ω = 2π/T = 71808 Hz

!  The peak at t1 has a height of V1 = 50 V !  Your graph may have a different voltage and period

Read voltage off the graph !  From the graph:

•  V0 = 70 V •  V1 = 50 V at t1 = 0.0007 s

!  Voltage in a damped oscillating circuit:

!  And each time the wave peaks, cos(ωt)=1, thus

!  Here: R = 57.6 Ohm and R/2L = 480.8 s-1

Chapter 30 4

V =V0e− Rt

2L cos ωt( )

V1 =V0e

−Rt12L , hence R =

−2L ln V1 V0( )t1

Determine the capacitance !  From the frequency ω, determine first the resonance

frequency ω0

!  And then determine the capacitance as

!  ω0 = 71810 Hz, very close to ω because the exponential decay is gradual

!  Thus C = 3.24*10-9 F = 3.24 nF

Chapter 30 5

ω = ω0

2 − R2L

⎛⎝⎜

⎞⎠⎟

2

thus ω0 = ω 2 + R2L

⎛⎝⎜

⎞⎠⎟

2

ω0 =

1LC

thus C = 1Lω0

2

Chapter 30 6

Determine the initial energy !  The initial energy in the capacitor is given by

!  This is also the total initial energy in the circuit. Each time the voltage in the graph is at a maximum, the energy in the capacitor is at a maximum.

!  This capacitor energy decays according to

!  The cos2 term is =1 because we only consider the points in time where all the energy is in the capacitor

U0 =

12

CV 2 = 7.94×10−6 J

UE (t)=U0e−Rt

L

Question 1 !  Calculate the energy in the circuit after a time of 12 periods.

Note that the curve passes through a grid intersection point. !  After 12 periods, t12 = 0.00105 s and

!  Thus the answer to the first question is 2.89E-6 J !  This makes sense: It is roughly 1/3 of the initial energy,

which would correspond to about 60% of the original voltage, which makes sense from the graph.

!  You will likely read off different values from your graph and will have a different inductance. But you should be able to follow the steps in here.

Chapter 30 7

UE (t)=U0e−Rt

L = 7.94×10−6e−961.6∗t12 J = 2.89×10−6 J

Question 2 !  Calculate the time required for 93% of the initial energy to

be dissipated. !  In order for 93% of the initial energy to be dissipated, we

need 7% of the initial energy remaining, or

!  Thus

!  This makes sense, it is a bit longer longer than t12

Chapter 30 8

U93(t93 )=U0e−Rt93

L where U93 = (1−0.93)U0

0.07 = e

−Rt93L and t93 = −

LR

ln0.07 = 0.00277s

Conclusions !  This is a complicated problem involving many steps !  You should know how to read numbers off the graph

•  Period, and converting it to ω •  You should be able to compute the capacitance C •  Voltage and its time dependence •  You should be able to compute R

!  You should understand how to calculate the initial energy U0

Chapter 30 9