Answer to Question #201914 in Mechanics | Relativity for fexypotato

Question

A 2.00 [kg] frictionless block is attached to an ideal spring of force constant 300. [N/m]. At t = 0 the

block has velocity -4.00 [m/s] and displacement 0.200 [m]. Determine the following:

A.The amplitude of the oscillation

B. The position-time and velocity-time functions of the spring-mass system

Answer 1

For a spring-mass system without any friction we have the following equation that relates the total energy of the system with the position and velocity of the oscillating mass:

$\frac{1}{2}kx^2+\frac{1}{2}mv^2=\frac{1}{2}kA^2\implies A = \sqrt{\frac{mv_{(t)}^2}{k} +x_{(t)}^{2}}$

Then, we substitute the velocity and position at the start with the spring constant and the mass to find the amplitude A:

$A = \sqrt{\frac{(2\,kg)(-4\,m/s)^2}{200\,N/m} +(0.2\,m)^{2}}=0.4472 \,m$

B) Then, the position-time and velocity-time functions of the spring-mass system would be

$x(t) = A \cos(\omega t + \phi)\\ \frac{d}{dt}x(t)=v(t) = -A\omega \sin(\omega t + \phi)$

with the constants:

1. A = 0.4472 m
2. $ω=\sqrt{\frac{k}{m}}=\sqrt{\frac{300\,N/m}{2\,kg}}=\sqrt{150}\,s^{-1} =12.247\,s^{-1}$
3. $\phi=cos^{-1}(\frac{x_{(0)}}{A})=63.434°$

Reference:

- Young, H. D., Freedman, R. A., & Ford, A. L. (2006). Sears and Zemansky's university physics (Vol. 1). Pearson education.