A 1.94 kg object on a frictionless horizontal track is attached to the end of a

Question

A 1.94 kg object on a frictionless horizontal
track is attached to the end of a horizontal
spring whose force constant is 4.73 N/m. The
object is displaced 3.04 m to the right from its
equilibrium position and then released, which
initiates simple harmonic motion.
What is the magnitude of the force acting
on the object 3.99 s after it is released?
Answer in units of N

i got the first part but what equation do i use to find the second part:

How many times does the object oscillate in 3.99 s? answer in units of turn?

Explanation / Answer

To get the answer to the second part, we use the period of a spring formula:

T = 2 ( m/k )

T = 2 ( 1.94 / 4.73 )

T = 4.024

This means that it takes 4.024 seconds to complete ONE FULL CYCLE.

However, it is asking for only 3.99 sec. Therefore, it hasn't completed one full cycle ( or turn ).

3.99/4.024 = .99155 TURNS

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