# Harmonic Oscillation Problem

PhysicS FAN
Homework Statement:
An object of mass m=1kg is executing a simple harmonic oscillation. During its oscillation it passes from a point A at which the speed of the object is u=1m/s and the force that its acting on it is F=2N. When the object passes form point B is moving with a speed u'=2m/s and an acting force of F'=1N. What's the frequency of the oscillation and what's the biggest distance from the equivalence position.
Relevant Equations:
T=2π/ω, ΔΚ= W, F=kd
First of all, I found a function of the distance of the object form the equivalence point in both cases. I got something like d=2d' where d is the distance at the first case and d' at the second. I did that because I wanted to find the frequency, and so first I need to find the period of oscillation. In order to find the period I need to know D ( T=2π √m/D ). I used energy theorems to get an other equation involving d and d' but I am stuck. I think that the problem is quite easy but I just can't get to the answer. Any Ideas?

## Answers and Replies

Science Advisor
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##F = ma## is one idea.

Science Advisor
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It looks tricky. Not easy. But, maybe if you look at the equations for SHM, you could find ##\cos(\omega t_A)## and ##\sin(\omega t_A)## etc.

That might help.

PhysicS FAN
It looks tricky. Not easy. But, maybe if you look at the equations for SHM, you could find ##\cos(\omega t_A)## and ##\sin(\omega t_A)## etc.

That might help.
Yes I'll try that

PhysicS FAN
Yes I'll try that
Well, I can't find anything that way. I think the best way to approach it is energetically since D theoritically could be found this way.

Science Advisor
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Well, I can't find anything that way. I think the best way to approach it is energetically since D theoritically could be found this way.

It definitely comes out that way. You could start with the relationship between the forces implies the same relationship between the accelerations at ##A## and ##B##. So:

##\cos(\omega t_A) = 2\cos(\omega t_B)##

The relationship between the velocities gives you another equation for the sines. From these you should be able to solve for all the relevant sines and cosines.

You still need the energy equation and what you've got already. But, getting the value of the sines and cosines is key.

PhysicS FAN
It definitely comes out that way. You could start with the relationship between the forces implies the same relationship between the accelerations at ##A## and ##B##. So:

##\cos(\omega t_A) = 2\cos(\omega t_B)##

The relationship between the velocities gives you another equation for the sines. From these you should be able to solve for all the relevant sines and cosines.

You still need the energy equation and what you've got already. But, getting the value of the sines and cosines is key.
Thanks a lot you've been really helpful!