Question

A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 6.0 cm at the moment the quake stops, and 7.0 s later it is 2.1 cm .

a.What is the time constant for the damping of the oscillation?

b.What was the amplitude of the oscillation 3.5 s after the quake stopped?

Answer 1

given

The amplitude of the vibration of the top of the lamppost is 6.0 cm

so x₁ = 6 cm

t₁ = 0

7.0 s later = t₂

it is 2.1 cm = x₂

a)

using equation

time constant = ( t₁ - t₂ ) / ln ( x₂/x₁ )

= ( 0 - 7 ) / ln ( 2.1 / 6 )

the time constant for the damping of the oscillation is = 6.6677 sec

b )

y = y₁ e^t/

= 2.1 x e^(3.5/6.6677)

y = 3.549 cm

the amplitude of the oscillation 3.5 s after the quake stopped is y = 0.03549 m