Question

An oscillator with a mass of 510 g and a period of 1.20 s has an amplitude that decreases by 1.30% during each complete oscillation.

Part A

If the initial amplitude is 13.6 cm , what will be the amplitude after 50.0 oscillations?

cm

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Part B

At what time will the energy be reduced to 51.0% of its initial value?

s

Answer 1

An oscillator with a mass of 510 g and a period of 1.20 s has an amplitude that decreases by 1.30% during each complete oscillation.

(A)
After 1 oscillation the amplitude is reduced by 1.3% = 0.987.
After 2 oscillations it will be reduced again by 0.987.
So after 50 oscillations it will be reduced by a factor 0.987^50

Final Amplitude A = Ao * 0.987^50
A = 13.6 * 0.987^50
A = 7.07 cm

(B)

As we know,
E = kA²
If Amplitude is reduced to 51.0% of its initial value !!
Amplitude is reduced to sqrt(0.51) of it's initial values.

Final Amplitude = 13.6 * sqrt(0.51) = 9.71

Now Let no of oscialltions needed to reduce amplitude to this value, be N
9.71 = 13.6 * 0.987^N
Taking log
log(9.71/13.6) = N * log(0.987)
N = -0.1463/-0.00568
N = 26