Simple Harmonic/Spring Question

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The discussion revolves around solving two problems related to simple harmonic motion and spring mechanics. For problem 4, participants clarify that the period of a simple harmonic oscillator can be calculated using T = 2π/ω, where ω is derived from k/m. In addressing problem 3, the focus shifts to visualizing the motion of two particles and understanding their displacement in relation to amplitude. Participants emphasize the importance of conceptualizing the motion rather than solely relying on equations. Overall, the conversation highlights the need for a deeper understanding of the physical principles behind harmonic motion.
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http://www.quantumninja.com/hw/random/34.JPG

I was hoping someone could help me with those two problems

I am not sure how to start 3. Here is what I have done with 4 so far...

\sum F= -kx-mg

k=245

\sum F= -kx-mg

\sum F= 34.3

F=ma
F=m x(double prime) where x= Acos(\omega t + \theta)

after that I am lost

Please help
thx
 
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HINT for 4.U don't need the differential equation.Nor the acceleration of gravity.

Daniel.
 
Are you doing this for a course in physics, or differential equations?

If it's for an elementary physics course, as dextercioby said, solving that differential equation is a lot more work than you need to do.

The period of a simple harmonic oscillator is given by

T = \frac{2 \pi}{\omega}

and \omega = \sqrt{ \frac{k}{m}}

=======================================================

As for #3, put your calculator away & visualize the motion of the particles. Slide your fingers around if necessary. Concentrate on the meanings of frequency, amplitude, displacement, and phase difference.

Think, what do they mean by "each time their displacement is half their amplitude". When does that happen?
 
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ok for number 4 according to what I believe you meant

is it just

\omega = \sqrt{490}

and therefore

T= \frac{2\pi}{\sqrt{490}}
 
For number 3 I have been looking at the equation

x=X_{m}cos(\omega t + \phi)

Am I going about this the right way. I know there should be a simple set here, but I am missing it
 
Stop looking at equations. What does this mean:
"They pass each other going in opposite directions each time their displacement is half their amplitude."

It says the frequencies and amplitudes are equal, right? Let's say you have two particles A and B and the period of each one is 4 seconds (just making up a number) and the amplitude is just the length of this:

================================​

Redraw that a few times & try marking on each one the positions of the two particles at each time for t=0, t=1, t=2, t=3, t=4 (seconds)
 
\frac{2\pi}{3}
 
I am getting \pi
 
Tom, it's hard to help you if I don't know how you're thinking about it.

Suppose the particles are oscillating with a frequency of .25 cycle/s

Suppose at time 0 two particles A and B are at opposite ends of the line:

T=0...A========================B


Where will they be at these times:


T=1s...========================

T=2s...========================

T=3s...========================

T=4s...========================
 
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