Question

An object oscillates with an angular frequency of 8.0 rad/s. At t = 0, the object is at x0 = 4 cm with an initial velocity v0 = -25 cm/s. Find the amplitude and the phase constant for the motion.

Answer 1

Answer:

$\phi = 0.66 rad$

$A = 5.06 cm$

Explanation:

We have here a simple harmonic motion, so the equation of the position in this motion is:  

$x(t)=Acos(\omega t+\phi)$ (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:

$v(t)=\frac{dx(t)}{dt}=-A\omega sin(\omega t+\phi)$ (2)

Let's evaluate (1) and (2) in t=0.

$x(0)=Acos(\phi)$ (3)

$v(0)=-A\omega sin(\phi)$ (4)

Dividing 4 by 3 we have:

$\frac{v(0)}{x(0)}=-\omega tan(\phi)$

$\phi = tan^{-1}(\frac{-v(0)}{\omega x(0)})$

$\phi = 0.66 rad$

Now, using (3) we can find the amplitude.

$A = \frac{x(0)}{cos(\phi)} = 5.06 cm$

I hope it helps!