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?
mass and volume are the answers you are looking for
Ithink the answer would be a
(a) T=6.07s
(b) [tex]x_{max|4.0s}=3.62cm[/tex]
Explanation:
For Part (a)
The initial amplitude is given as A=7.0 cm
Apply the equation [tex]x_{max}(t)=Ae^{-t/T}[/tex] with [tex]x_{max}(7.6s)=2.0cm[/tex] we have:
[tex]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[/tex]
For Part (b)
Apply [tex]x_{max}(t)=Ae^{-t/T}[/tex] with t=4.0s and T=6.07s we have
[tex]x_{max}(t)=Ae^{-t/T}\\x_{max}(4.0s)=(7.0cm)e^{-4.0/6.07}\\x_{max|4.0s}=3.62cm[/tex]