Answer to Question #167429 in Quantum Mechanics for Kipkirui Emmanuel

Question

A block attached to a spring is released at t =0 with the spring extended.the period of oscillation is 0.61s.At t = 0.05s the velocity is v=-96.4cm/a.(1)what is the amplitude? (2)what is the total energy?

Answer 1

$x=Acos \omega t=Acos \frac{2\pi t}{T},$

$v=-\frac{2\pi A}{T}sin\frac{2\pi t}{T}, \implies$

$A=-\frac{vT}{2\pi sin\frac{2\pi t}{T}},$

$A=-\frac{-0.964\cdot 0.61}{6.28\cdot sin\frac{6.28\cdot 0.05}{0.61}}=0.19~m,$

$E=\frac{kA^2}{2}=\frac{\omega ^2 m A^2}{2}=\frac{2\pi ^2mA^2}{T^2}=\frac{2\cdot 3.14^2\cdot m\cdot 0.19^2}{0.61^2}=1.91m~(J),~$ where $m$ is the mass of a block.