1. Damp Oscillation
Tiffany Yang
18422148

2. •Supposed that a damped mass-spring oscillator
loses 4% of its energy during each cycle. If the
period is 2.0 s, the mass is 0.20 kg, and the initial
amplitude is 0.30m
Question 1: What is the value of the
damping constant

3. Explanation
• Damped oscillator oscillates at a lower frequency, and the
damped amplitude decreases over time. We combined the
amplitude and a exponential term to express a relationship
between amplitude and time passes.
• Key equation: A(t)=Ae^(-kt/2m)

4. Calculation:• the initial amplitude is 0.3
• Amplitude of the oscillator
after 1 oscillation: 0.3*(1-
4%)=0.288
• We know that T=2.0, and
mass is= 0.20
• We can then plug in all the
values into the equation:
A(t)=Ae^(-k/m)
0.3*e^(-kt/2m)=0.288
e^(-kt/2m)=0.94
ln94=-kt/2m
k=-2m/t*ln(0.95)
k=-0.20*2/2.0*ln(0.94)
k=0.0124kg/s

5. Question 2: How many cycles elapsed before
the amplitude reduces to 0.20m?
• We can also apply the key equation A(t)=Ae^(-kt/2m) in this question.
• We can use this equation to calculate the time elapsed.
Calculation of t:
A(t)=Ae^(-kt/(2m))
A(t)/A=e^(-kt/(2m))
ln(0.2/0.3)=-kt/(2m)
ln(3/2)=kt/(2m)
t=ln(2/3)*2m/k
t=2*0.20/(0.0124)*ln(2/3)
t=13.08s

6. Number of cycle=time/period
•t/T=13.08/2.0
•t/T=6.54
•Therefore, around 6.54cycles elapsed before
amplitude reaches to 0.20

7. Extended Question(no calculation is
required):
• 1. Would it take the same number of circles to reduce
the amplitude from 0.2 to 0.1?
• 2. How would the number of circles change if you
decrease the damping constant.

8. Answers:
• 1. No. According to the equation A(t)=Ae^(-kt/2m), the amplitude
decreases exponentially not linear. So it takes different amount of
time to reduce the amplitude from 0.2 to 0.1.
• 2. Number of circles will increase. Damping constant measures the
strength of the drag force. So as you decreases the constant, the drag
force decreases. Number of circles oscillate increases. Also, you can
observe the relationship from the equation A(t)=Ae^(-kt/2m) that t
and k are inversely proportional.