Answer:
[tex] \phi = 0.66 rad [/tex]
[tex] A = 5.06 cm [/tex]
Explanation:
We have here a simple harmonic motion, so the equation of the position in this motion is:
[tex] x(t)=Acos(\omega t+\phi) [/tex] (1)
A: Amplitude
ω: angular frequency
φ: phase constant
If we take the derivative of x with respect to t from (1), we can find the velocity equation of this motion:
[tex] v(t)=\frac{dx(t)}{dt}=-A\omega sin(\omega t+\phi) [/tex] (2)
Let's evaluate (1) and (2) in t=0.
[tex] x(0)=Acos(\phi) [/tex] (3)
[tex] v(0)=-A\omega sin(\phi) [/tex] (4)
Dividing 4 by 3 we have:
[tex] \frac{v(0)}{x(0)}=-\omega tan(\phi) [/tex]
[tex] \phi = tan^{-1}(\frac{-v(0)}{\omega x(0)}) [/tex]
[tex] \phi = 0.66 rad [/tex]
Now, using (3) we can find the amplitude.
[tex] A = \frac{x(0)}{cos(\phi)} = 5.06 cm [/tex]
I hope it helps!
