Answer to Question #257617 in Mechanics | Relativity for Waseem

Question

A block attached to a spring is made to oscillate with initial amplitude of 8.0 cm. After 2.2 minutes, the amplitude decreases to 5.0 cm. Calculate (i) the time when the amplitude becomes 2.0 cm, and (ii) the value of damping constant  for this mot

Answer 1

$A_0 = 8cm = 0.08m$

$A_1 = 5cm = 0.05m$

$t_1 = 2.2min = 132s$

$A_2 = 2cm = 0.02m$

$A _1= A_0e^{-\gamma t_1}$

$\gamma=-\frac{1}{t_1}*\ln\frac{A_1}{A_0}= -\frac{1}{132}*\ln \frac{0.05}{0.08}=3.56*10^{-3}$

$A _2= A_0e^{-\gamma t_2}$

$t_2=-\frac{1}{\gamma}*\ln\frac{A_2}{A_0}= -\frac{1}{3.56*10^{-3}}*\ln \frac{0.02}{0.08}\approx389s$

$\text{Answer:}$

$\text{(i) }t_2=389s$

$\text{(ii) }\gamma=3.56*10^{-3}$