Answer to Question #314338 in Mechanics | Relativity for Sanghamitra bemal

Question

A block attached to a spring is made oscillate with initial amplitude of 8.0 cm after 2.2 minutes, the amplitude decrease to 5.0 cm calculate

Answer 1

$A_0 = 8cm = 0.08m$

$A_1 = 5cm = 0.05m$

$t_1 = 2.2min = 132s$

$A_2 = 2cm = 0.02m$

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

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

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

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

$\text{Answer:}$

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

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