Harmonic oscillator problem

[mkerikss]

Homework Statement

A particle with with the mass of m is attached to a spring (with no mass, spring constant k, length l) which is attached to a wall. The particle is moving with no friction along the x-axis.
a) Write the particles motion equation, and find the general solution to the motion using a test-method.
b) Using the solution, determine which are the classic turning points.
c)What is the work done by the spring-force from x₀ to x, and show that the force is conservative. Write the spring-force' potential energy U's expression.
d) I have solved it and it isn't needed anymore.
e)Write the harmonic oscillator's principle of conservation of energy.
f)Show usin the solution to a), that the particle's total energy is constant. What is the constant.
g)Write the general solution to the motion equation for a harmonic oscillator from the principle of conservation of energy.
Integrate: [tex]\int((a+bx+cx²)^-1/2)[/tex]=-(-c)^-1/2arcsin((2cx+b)/\sqrt{b²-4ac} , c<0, b²-4ac>0
(I have no idea what this question's all about so I may very well have translated it incorrectly)
h) Determine the total energy in g) and use it in f)
i) Determine the total energy in e) using the principle of conservation's classic turning points. Did you get the same result as in b). Hint: The mechanical energy\geq0

Homework Equations

The Attempt at a Solution

NII gives -kx=ma
Eq.1: d²x/dt²+kx/m=0

I used x(t)=Asin(rt+\theta)
x''(t)=Ar²(-sin(rt+$$\theta$$))

I put these into eq.1 and got r=(n$$\pi$$-$$\theta$$/t and r=$$\sqrt{k/m}$$

I don't really know what to do with the first answer, but the second gives me the answer we're supposed to get.

Eq.2: x(t)=Asin(\omegat+\theta)

b) Not sure about this one, as I'm unsure what theses points are, but someone told me that they are the points when the amplitude +-A. Is this correct?

c)W=\intF*ds=-kx²/2+kx_{0}²/2 (the integrat is from x_{}0 to x)

W(x₀ to x)=U(x₀)-U(x)

U(x)=-W(x₀ to x)=kx²/2-kx_{0}²/2=kx²/2
because U(x₀)=0, and x₀=0

d)I have solved it and I don't need it in the rest of the problem so I won't write it here.

e)E=mx²/2+kx²/2
The second term is the potential energy, but I don't know where the first term comes from. It must be the mechanical energy but why does it look like that?

f) I have tried to put Eq.2 into the equation above (in e)) to get a constant but that doesn't work, and now I don't know what to do. Help me please!

g,h,i) I really don't have any idea how to solve these.

Edit: Why does this text look sp weird. I hate computers

 

[Delphi51]

a) I don't see where you got the first value for r. Looks like r^2 = k/m which gives only the second, correct value.
b) Yes, I think the turning points are where the mass reverses direction, which is when x = A.
c) looks okay
e)E=mx2/2+kx2/2
Surely this should be kinetic + potential energy so the first term would be .5*m(v)^2 or .5*m*(x')^2
f) With the above, you probably won't have any trouble with f. You'll have to use the r^2 = k/m to get it to work out to some constants times cos^2 + sin^2 which equals 1.

I don't know how to do the last parts, either. Maybe you are supposed to solve .5*m(x')^2 + .5*k*x^2 = constant?
Would that integral appear in the solution when you try to clear out the x' derivative?

Have to go now, but hope to see you write the solution here!

 

[tomcue]

I understand the difficulty in solving complex problems such as the harmonic oscillator problem. However, let me try to clarify some of the concepts and steps involved in solving this problem.

a) The particle's motion equation is given by Newton's Second Law, F=ma, which in this case becomes -kx=ma. This can also be written as m(d2x/dt2)+kx=0. To find the general solution, we use the method of undetermined coefficients, where we assume a solution of the form x(t)=Asin(rt+\theta). Substituting this into the equation and solving for r, we get two possible values for r, one of which is the correct solution: r=\sqrt{k/m}. Substituting this value into the assumed solution gives us the general solution x(t)=Asin(\sqrt{k/m}t+\theta).

b) The classic turning points are the points where the particle's velocity becomes zero, i.e. the points where the particle changes direction. These points can be found by setting the velocity equation equal to zero and solving for x. In this case, the turning points occur at x=\pm A, where A is the amplitude of the oscillation.

c) The work done by the spring force is given by the integral of the force with respect to displacement, W=\int Fdx=-kx2/2+kx_{0}2/2. This shows that the work done is dependent only on the initial and final positions of the particle, and not on the path taken. This is a characteristic of conservative forces. The potential energy of the spring force can be written as U(x)=kx2/2.

e) The harmonic oscillator's principle of conservation of energy states that the total mechanical energy of the system (kinetic energy + potential energy) remains constant throughout the motion. In this case, the total energy is given by E=mx2/2+kx2/2.

f) To show that the particle's total energy is constant, we can substitute the general solution from part a) into the expression for total energy (from part e)) and see that it remains constant. The constant value is the sum of the kinetic and potential energies, E=mx2/2+kx2/2.

g) The general solution to the motion equation for a harmonic oscillator can also be derived from the principle of conservation of energy. We can write