Amachine part is vibrating along the x-axis in simple harmonic motion with a period of 0.27 s and a range (from the maximum in one direction to the maximum in the other) of 3.0 cm. at time t = 0 it is at its central position and moving in the +x direction. what is its position when t = 55 s?

Explanation:

The general equation for position of Simple harmonic motion is given as:

         ........(1)

where,

x = Position of the wave

A = Amplitude of the wave

ω = Angular velocity

t = time

In this case, the amplitude is just half the range,

thus,

 (Given range = 3cm)

A = 1.5 cm  

Now, The angular velocity is given as:

Where, T = time period of the wave =0.27s (given)

or

so, at time t = 55 s, the equation (1) becomes as:

on solving the above equation we get,

here the negative sign depicts the position in the opposite direction of +x

what is the question?

f=ma a/f/m a= 480/60 a = 8m/s/s that is quick wow