Answer :
Answer:
[tex]x(t) = (1.667\,m)\cdot \cos [(3\,\frac{rad}{s} )\cdot t -\frac{\pi}{2} ][/tex]
Step-by-step explanation:
The statement of the problem shows a case of simple harmonic movement, whose expressions for position is:
[tex]x(t) = A \cdot \cos \left(\omega\cdot t + \phi \right)[/tex]
An expression for velocity is found by deriving the previous one:
[tex]v(t) = -\omega \cdot A \cdot \left( \omega\cdot t + \phi\right)[/tex]
The angular frequency is:
[tex]\omega = \sqrt{\frac{F_{k}}{\Delta x\cdot m} }[/tex]
[tex]\omega = \sqrt{\frac{360\,N}{(20\,kg)\cdot (2\,m)} }[/tex]
[tex]\omega = 3\,\frac{rad}{s}[/tex]
Amplitude and phase angle are found by replacing and simplifying motion formulas:
[tex]A \cdot \cos\phi = 0\,m\\-(3\,\frac{rad}{s} )\cdot A\cdot \sin \phi = 5\,\frac{m}{s}[/tex]
[tex]\phi = -\frac{\pi}{2}[/tex]
[tex]A = 1.667\,m[/tex]
The equation of motion is:
[tex]x(t) = (1.667\,m)\cdot \cos [(3\,\frac{rad}{s} )\cdot t -\frac{\pi}{2} ][/tex]