Differential Equations - Simple Harmonic Oscillation

[twiztidmxcn]

Homework Statement

Consider y''(t)+(k/m)*y = 0 for simple harmonic oscillation

A) Under what conditions on Beta is y(t)=cos(Beta*t) a solution?

B) What is the period of this solution?

C) Sketch the solution curve in the yv-plane associated with this solution (Hint: y^2 + (v/Beta)^2)

For A, I had:

dy/dt = v

dv/dt = -(k/m)*y

Found y' and y''

y'(t) = -Beta*sin(Beta*t) y''(t)=-Beta^2*cos(Beta*t)

Plugged those into the given equation and found Beta = +/- sqrt(k/m)

y(t) = cos (+/-sqrt(k/m)*t) -> answers for A

For B, I found the period to be T = 2*pi / Beta

For C, I found that y^2 + (v/Beta)^2 = cos^2(Beta*t) + Beta^2*sin^2(Beta*t)/Beta^2

My problem now is drawing this. I think, from remembering equations of shapes, that this is an ellipse, stretching in the v direction.

I am not sure, but I think the max and min points of the ellipse are:
y = 0, v = +/- sqrt(k/m)
v = 0, y = +/- 1

I also think that the direction of the field is counter-clockwise.

I don't know if part B and C were done totally right and am a bit confused about finding the direction of the phase field/solution.

-twiztidmxcn

 

[HallsofIvy]

Homework Statement

Consider y''(t)+(k/m)*y = 0 for simple harmonic oscillation

A) Under what conditions on Beta is y(t)=cos(Beta*t) a solution?

B) What is the period of this solution?

C) Sketch the solution curve in the yv-plane associated with this solution (Hint: y^2 + (v/Beta)^2)

For A, I had:

dy/dt = v

dv/dt = -(k/m)*y

Found y' and y''

y'(t) = -Beta*sin(Beta*t) y''(t)=-Beta^2*cos(Beta*t)

Plugged those into the given equation and found Beta = +/- sqrt(k/m)

y(t) = cos (+/-sqrt(k/m)*t) -> answers for A

Pretty good- since cosine is an evern function, you don't need the "+/-", both signs give the same function, but, yes, Beta= sqrt(k/m).

For B, I found the period to be T = 2*pi / Beta

Wouldn't it be better to replace Beta by the value you found in (a)?

For C, I found that y^2 + (v/Beta)^2 = cos^2(Beta*t) + Beta^2*sin^2(Beta*t)/Beta^2

equals what?

My problem now is drawing this. I think, from remembering equations of shapes, that this is an ellipse, stretching in the v direction.

I am not sure, but I think the max and min points of the ellipse are:
y = 0, v = +/- sqrt(k/m)
v = 0, y = +/- 1

Yes, that is corret.

I also think that the direction of the field is counter-clockwise.

WHY do you think that? Suppose you were at (0, sqrt(k/m)) on this ellipse. Since the second component, the rate of change in y, is positive, is y increasing or decreasing? Which direction does that tell you?

I don't know if part B and C were done totally right and am a bit confused about finding the direction of the phase field/solution.

-twiztidmxcn