Is this a simple harmonic question or no?

[tarletontexan]

Is this a simple harmonic question or no??

Homework Statement

a particle moves on the x-axis under the influence of a force, proportional to its distance from the origin, attracting it toward the origin. The particle starts from rest @x=10cm and reaches x=5cm in 2 seconds for the first time. What is the position t at any time after it starts movin, the amplitude, period and frequency of the motion.

Homework Equations

T=1/(omega), x=Acos(kx-omega t)

The Attempt at a Solution

I believe the amplitude of the system to be 10 cm simply because that is what the particle starts at, the equation of motion seems to be simple harmonic since it will go and come in a rhythmic pattern. Other than that i don't quite understand how to figure the rest of the problem

 

[Pengwuino]

At first glance this problem isn't worded well. You need to know the equilibrium position or the spring constant before you can solve this thing.

 

[tarletontexan]

there is no spring at all mentioned in this problem...only that a force proportional to its distance from the origin is applied to the particle.

 

[fzero]

It's a harmonic oscillator based on the force law. You're given the velocity at two values of the position, which is unorthodox, but I believe enough to solve for the frequency and amplitude.

 

[diazona]

there is no spring at all mentioned in this problem...only that a force proportional to its distance from the origin is applied to the particle.

It doesn't matter whether there's actually a spring there or not. What's important is that the force is proportional to displacement from the equilibrium position and attracts the particle toward equilibrium. A spring is just one example of a system that produces this kind of force, but it's the most common one, so we call the proportionality constant between force and displacement the "spring constant" even when there is no actual spring involved.

I believe the amplitude of the system to be 10 cm simply because that is what the particle starts at,

You're right that the amplitude is 10 cm, but that's not just because the particle starts at that point. The particle's speed has to be zero when it is at its amplitude.

the equation of motion seems to be simple harmonic since it will go and come in a rhythmic pattern.

As fzero said, the reason it is simple harmonic is because the force is attractive and proportional to displacement. That's all you need to know that you're dealing with a simple harmonic oscillator.

What else do you know about simple harmonic oscillators? Hint: the equation

x=Acos(kx-omega t)

is almost correct. If you fix it, you should be able to use it to find all the other quantities you're asked for.