Question

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At t=0 the block has velocity -4.00 m/s and displacement +0.200 m.

Find:
(a) the amplitude and (b) the phase angle.
c) Write an equation for the position as a function of time.
Assume x(t) in meters and t in seconds

Answer 1

0.38296 m

1.02132 rad

`x(t)=0.38296cos(12.24744 t+1.02132)`

Step-by-step explanation:

Angular frequency is given by

`\omega=\sqrt{(k)/(m)}\\\Rightarrow \omega=\sqrt{(300)/(2)}\\\Rightarrow \omega=12.24744\ rad/s`

Displacement is given by

`x(t)=Acos(\omega t+\psi)\\\Rightarrow 0.2=Acos(12.24744* 0+\psi)\\\Rightarrow 0.2=Acos\psi\\\Rightarrow A=(0.2)/(cos\psi)`

Velocity is given by

`v=-A\omega sin(\omega t+\psi)\\\Rightarrow -4=(0.2)/(cos\psi)12.24744sin(12.24744* 0+\psi)\\\Rightarrow 4=(1)/(cos\psi)2.449488sin(\psi)\\\Rightarrow tan\psi=(4)/(2.449488)\\\Rightarrow \psi=tan^(-1)(4)/(2.449488)\\\Rightarrow \psi=1.02132\ rad`

The phase angle is 1.02132 rad

`A=(0.2)/(cos\psi)\\\Rightarrow A=(0.2)/(cos1.02132)\\\Rightarrow A=0.38296\ m`

The amplitude is 0.38296 m

The equation is given by

`x(t)=Acos(\omega t+\psi)\\\Rightarrow x(t)=0.38296cos(12.24744 t+1.02132)`

The equation is
`x(t)=0.38296cos(12.24744 t+1.02132)`