Oscillation Problem: Find Period and Potential Energy

In summary: Yes, that's correct. The work done by a force can be divided into kinetic energy and potential energy. The potential energy is the energy that is stored in the position or configuration of an object, while the kinetic energy is the energy of motion. The potential energy can be calculated by integrating the force over a certain distance. In this case, the potential energy is given by Ep=kx^2/2, where k is the spring constant and x is the displacement from the equilibrium position.
  • #1
Misheel
6
0

Homework Statement


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.

Homework Equations


F=ma

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
2*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)) ?

and also assuming that in SHORT oscillation :
it is almost like a spring :
F=-12*pi*x=-kx
k=12*pi

potential energy is Ep= k*x^2/2=6*pi*x^2 ?
 
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  • #2
Misheel said:

Homework Statement


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.

Homework Equations


F=ma

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
2*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)) ?
This is fine.

and also assuming that in SHORT oscillation :
it is almost like a spring :
F=-12*pi*x=-kx
k=12*pi

potential energy is Ep= k*x^2/2=6*pi*x^2 ?
I think the problem is looking for the potential corresponding to the original force, not the approximation.
 
  • #3
Thank You for your reply, Vela

umm...then, i have no other ideas other than using FORCE to solve this problem :PP 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 ? :P

Thanks
 
  • #4
How are potential energy and force related in general?
 
  • #5
vela said:
How are potential energy and force related in general?

Maybe the work done by force is divided into object's konetik and potential energy ?
 

1. What is an oscillation problem?

An oscillation problem is a type of physics problem that involves finding the period and potential energy of a system that is undergoing oscillatory motion. This type of motion refers to a back-and-forth movement that repeats itself over a certain period of time.

2. How do you find the period of an oscillation problem?

The period of an oscillation problem can be found by using the formula T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. This formula applies to simple harmonic motion, which is a type of oscillatory motion characterized by a constant amplitude and frequency.

3. What is potential energy in an oscillation problem?

Potential energy in an oscillation problem refers to the energy stored in a system due to its position or configuration. In the case of oscillatory motion, potential energy is usually associated with the position of the object in relation to the equilibrium point. As the object moves away from the equilibrium point, its potential energy increases, and as it moves towards the equilibrium point, its potential energy decreases.

4. How does the amplitude affect the period in an oscillation problem?

The amplitude, which is the maximum displacement of the object from its equilibrium point, does not affect the period in an oscillation problem. The period is only affected by the mass and spring constant of the system. However, the amplitude does affect the maximum potential energy of the system, as a larger amplitude results in a larger potential energy.

5. What are some real-life examples of oscillation problems?

Some real-life examples of oscillation problems include the swinging of a pendulum, the back-and-forth motion of a spring, and the motion of a mass attached to a vertical spring. Other examples include the vibrations of a guitar string, the motion of a swing, and the movement of a diving board after a diver jumps off of it.

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