Question

A small earthquake starts a lamppost vibrating back and forth. The amplitude of the vibration of the top of the lamppost is 7.0 cm at the moment the quake stops, and 7.6 s later it is 2.0 cm.a. What is the time constant for the damping of the oscillation? b. What was the amplitude of the oscillation 4.0 s after the quake stopped?

Answer 1

Answer:

(a) T=6.07s

(b) $x_{max|4.0s}=3.62cm$

Explanation:

For Part (a)

The initial amplitude is given as A=7.0 cm

Apply the equation $x_{max}(t)=Ae^{-t/T}$ with  $x_{max}(7.6s)=2.0cm$ we have:

$x_{max}(t)=Ae^{-t/T}\\2.0cm=(7.0cm)e^{-7.6s/T}\\T=-\frac{7.6s}{ln(\frac{2.0cm}{7.0cm} )} \\T=6.07s$

For Part (b)

Apply $x_{max}(t)=Ae^{-t/T}$  with t=4.0s and T=6.07s we have

$x_{max}(t)=Ae^{-t/T}\\x_{max}(4.0s)=(7.0cm)e^{-4.0/6.07}\\x_{max|4.0s}=3.62cm$