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REY [17]
3 years ago
12

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.
Physics
1 answer:
KatRina [158]3 years ago
6 0

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!                            

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