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 Periodic and Circular Motion Periodic and Circular Motion Physics Help Forum Nov 4th 2014, 08:13 AM #1 Junior Member   Join Date: Nov 2014 Posts: 2 Oscillation problem 1. The problem statement, all variables and given/known data mass "m" object is located near the origin of the coordinate system. External force was exerted on the object depending on the coordinate by the formula of Fx=-4*sin(3*pi*x). Find the short oscillation period, and the potential energy depending on the coordinate. 2. Relevant equations F=ma 3. The attempt at a solution I am not really sure what is the short oscillation period... but since there is only 1 force: F=Fx=ma -4sin(3*pi*x)=m*(d^2*x)/dt^2 and assuming that the object will move very little (because it's said to be SHORT osccilation period ?) sin(3*pi*x) is approximately 3*pi*x . and since -ω2*x=(d^2*x)/dt^2 is the formula of the harmonic oscillation : we find ω=12*pi/m and the T=2*pi*√(m/(12*pi)) ? then to find potential energy: umm..., i have no other ideas other than using VIOLENCE(hihi) to solve this problem :P which is : F=-kx Fx=F=-4*sin(3*pi*x)=-kx from this we find k=4*sin(3*pi*x)/x so Ep=kx^2/2=2*sin(3*pi*x)*x ??? is it right ?    Nov 4th 2014, 10:43 AM   #2
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 Originally Posted by MishkaMN and since -ω2*x=(d^2*x)/dt^2 is the formula of the harmonic oscillation : we find ω=12*pi/m
You mean ω^2=12*pi/m

 Originally Posted by MishkaMN then to find potential energy: umm..., i have no other ideas other than using VIOLENCE(hihi) to solve this problem :P which is : F=-kx Fx=F=-4*sin(3*pi*x)=-kx from this
I would leave it as F=-4 sin(3*pi*x), and perform the integral:

W(x) = -PE(x) = integral from s=0 to s=x of 4*sin(3*pi*s)ds

Last edited by ChipB; Nov 4th 2014 at 10:46 AM.   Nov 4th 2014, 10:44 PM   #3
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 You mean ω^2=12*pi/m I would leave it as F=-4 sin(3*pi*x), and perform the integral: W(x) = -PE(x) = integral from s=0 to s=x of 4*sin(3*pi*s)ds

yeah it was w^2 sorry didnt see that

But the work done by the force is divided into not only potential energy but also kinetik energy right ?
where is the kinetik energy ?

Last edited by MishkaMN; Nov 4th 2014 at 10:51 PM.   Nov 5th 2014, 01:01 AM #4 Physics Team   Join Date: Feb 2009 Posts: 1,425 If by short oscillation, they mean a small value of x then sinx = x and you can approximate the oscillation as simple harmonic as Fx = - 4*pi*x. In this case 4pi = omega^2 and the rest follows. In S.H.M. the total energy is a sum of kinetic and potential, so max P.E. at the ends (when most work is done) and zero K.E. At the centre P.E. = 0 since x = 0 and K.E. is maximum.   Nov 5th 2014, 04:54 AM   #5
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 Originally Posted by MishkaMN But the work done by the force is divided into not only potential energy but also kinetik energy right ? where is the kinetik energy ?
If you apply force F per the given equation the spring will compress to position 'x', and at that point it has velocity =0, and consequently KE=0.   Nov 11th 2014, 02:20 AM #6 Junior Member   Join Date: Nov 2014 Posts: 2 What are oscillations and types of oscillations? Let me give you a simple explanation of What are oscillations and types of oscillations? This will help you understand the problem. Click on the link http://bit.ly/1sywpcO For more such Physics Lesson Videos, Subscribe my YouTube Channel Click here http://bit.ly/1zPAxxG  Tags force, oscillation, potential energy, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Physics Forum Discussions Thread Thread Starter Forum Replies Last Post gregade Advanced Mechanics 1 Jun 27th 2012 10:10 AM HeatherN Waves and Sound 5 Sep 6th 2010 07:08 AM Bluekitten Advanced Mechanics 1 Nov 12th 2009 11:00 PM taichi2910 Kinematics and Dynamics 2 Jun 3rd 2009 04:15 PM dsptl Advanced Mechanics 1 Nov 9th 2008 01:57 PM