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## Homework Statement

The point of the needle of a sewing machine moves in SHM along the x-axis with a frequency of 2.5 Hz. At t=0 its position and velocity components are +1.1 cm and -15 cm/s, respectively.

(a) Find the acceleration component of the needle at t=0

(b) write an equation giving the position at the point as a function of time.

I just want to state that I already know the answers, my question is concerning the value of the Amplitude in both (a) and (b).

My first question:

So for part (a) the answer for acceleration is -2.71 m/s

^{2}. This is found by using the equation

a(t) = -ω

^{2}A = -2.71 m/s

^{2}

where ω = 5π and A = 0.011 m. Why is the amplitude 0.011m? In the question it states that at t=0 its at a certain position , 1.1cm, with a negative velocity, if it has a velocity then its not at rest, and if its not at rest that means its not at its max position away from the equilibrium point, which from what I understand is defined as the Amplitude. So why is 1.1 cm used as the amplitude in this case?

My 2nd question

In part (b) the answer for the position equation is:

x(t) = (0.0146)cos[(15.7 rad/s)t + 0.715 rad]

So in this case A = 0.0146. This is found by using the equation

A = [x

_{o}

^{2}+ v

_{o}

^{2}/ω

^{2}]

^{½}

So why does this have a different Amplitude from in part (a). Should the value for A be the same in both cases?