Aparticle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 n/m. at the moment t = 0, the particle has its maximum speed of 20.0 m/s and is moving to the left. a) determine the particleã¢â‚¬â„¢s equation of motion, specifying its position as a function of time. b) where in the motion is the potential energy three times the kinetic energy? c) find the minimum time interval required for the particle to move from x = 0 to x = 1.00 m. d) find the length of a simple pendulum with the same period.

a)

The equation which gives the position of a simple harmonic oscillator is:

where

A is the amplitude

is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

The amplitude, A, can be found from the maximum velocity of the spring:

So, the equation of motion is

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

While the kinetic energy is given by:

The velocity as a function of time t is:

The problem asks as the time t at which U=3K, so we have:

However, , so we have

with two solutions:

c) 3 seconds.

When x=0, the equation of motion is:

so, t=0.

When x=1.00 m, the equation of motion is:

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

The period of a pendulum is:

where L is the length of the pendulum. By using T=0.628 s, we find

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