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A simple question on simple harmonic motion

[tex]F_s=-kx=ma, a=\frac{-kx}{m}[/tex]

We now need to find an equation that satisfies: [tex]\frac{d^2x}{d^2t}=\frac{kx}{m}[/tex]

We get (letting w^2=k/m): [tex] x(t)=A*Cos(wt+\phi)[/tex]

The time it takes for one cycle is:[tex]T_1=2 \pi\sqrt{\frac{m}{k}[/tex]

Now, lets solve for T again using a different method.

[tex] F_{ave}=\frac{1}{x}\int{(kx)}dx=\frac{kx}{2}[/tex]
[tex]a_{ave}=\frac{kx}{2m}[/tex]

So, we get: [tex]x_f=x_i+\frac{at^2}{2}--->T^2=\frac{4m}{k}--->T=2\sqrt{\frac{2m}{k}}[/tex]

We multiply our new T by 4 to get the time over a complete cycle: [tex]T_2=4*2\sqrt{\frac{2m}{k}}[/tex]

[tex]T_1/=T_2[/tex]

? ? ? ? ? ? ? ? ?
 

Dick

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Your first method makes sense if you change the ODE to x''=(-k/m)x. Your 'second method' doesn't make any sense at all. The average force and acceleration over a full cycle of cos(w*t) are zero. What does F_ave=k*x/2 mean? x is variable.
 
No, F_ave is over 1/4 a cycle, from x=A to x=0. F_ave=kx/2 is kx integrated over x and then divided by the length it was integrated over.
 
...oopsie.

I see now what I did. My x=0 point is not the point where the spring is all the way compressed, my x=0 is where the spring is only halfway compressed,
 
Hmm..still not really making sense to me.

I must be doing something basic wrong. Okay, lets just ignore the first method. If I had a block attached to an unstretched string, and then I stretched it to x=A, how would I go about solving for T. When T is the time it takes the spring to get the block to x=-A.
 

Dick

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You do it the same way everyone else does it. You solve the differential equation for the spring with your initial conditions. You shouldn't ignore the first method. It's the second method that's wrong.
 
Well, even if you solve that differential equation, you'd end up with the following solution (as i had in another thread I wrote myself):

[tex]x(t)=C_{1}cos(\omega_{0}t)+C_{2}sin(\omega_{0}t)[/tex]

I never understood how to go from that to the solution you see in physics textbooks. Also, your differential equation you wrote in the first part is missing a negative sign in front of the kx.

Anyways, the integral you took in the second method would give you the work divided by the distance. But I guess that does equal force?
 
Last edited:
So, we get: [tex]x_f=x_i+\frac{at^2}{2}--->T^2=\frac{4m}{k}--->T=2\sqrt{\frac{2m}{k}}[/tex]
It looks like you're trying to apply the displacement equation for constant acceleration here. Additionally, it looks like you solved for T, in your own equation, incorrectly. That might be a little beside the point, though.
 
It looks like you're trying to apply the displacement equation for constant acceleration here. Additionally, it looks like you solved for T, in your own equation, incorrectly. That might be a little beside the point, though.
[tex]x_f=x_i+V_iT+\frac{at^2}{2}[/tex], where [tex]a=\frac{-kx}{2m}, x_f=-A,x_i=+A[/tex]

[tex]-2A=\frac{\frac{-kx}{2m}*t^2}{2}=-2A*\frac{2}{\frac{-kx}{2m}}

=T^2=\frac{8m}{k}, T=(\frac{8m}{k})^(1/2)[/tex]

What did I do wrong with the algebra?

As for the physics, your saying if I have a non constant a, I can't integrate it,divide by the domain I integrated over, and then use that in the constant a kinematic equations?

This is all self study at this point, I am starting college in august, but thought I should go over this stuff before then, as my high school was...less then stellar.
 

cepheid

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Now, lets solve for T again using a different method.

[tex] F_{ave}=\frac{1}{x}\int{(kx)}dx=\frac{kx}{2}[/tex]
Umm no. This definition of the average value of a function is wrong. You divide the integral by some NUMBER that represents the interval over which you averaged. Dividing by x makes no sense, because x is the variable of integration (as has been mentioned). IF you are averaging over a length A as you stated in your second post, then it should be:

[tex] F_{\textrm{ave}} = \frac{1}{A}\int_0^A kx\, dx [/tex]


EDIT: Oh yeah, and then there's the main point, which is, how does calculating this *help* you? The acceleration is NOT constant, so pretending that it is and that it is equal to its average is unlikely to yield the correct answer.
 

cepheid

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Well, even if you solve that differential equation, you'd end up with the following solution (as i had in another thread I wrote myself):

[tex]x(t)=C_{1}cos(\omega_{0}t)+C_{2}sin(\omega_{0}t)[/tex]

I never understood how to go from that to the solution you see in physics textbooks.
What solution "in textbooks" are you referring to?
 
[tex]x_f=x_i+V_iT+\frac{at^2}{2}[/tex], where [tex]a=\frac{-kx}{2m}, x_f=-A,x_i=+A[/tex]

[tex]-2A=\frac{\frac{-kx}{2m}*t^2}{2}=-2A*\frac{2}{\frac{-kx}{2m}}

=T^2=\frac{8m}{k}, T=(\frac{8m}{k})^(1/2)[/tex]

What did I do wrong with the algebra?

As for the physics, your saying if I have a non constant a, I can't integrate it,divide by the domain I integrated over, and then use that in the constant a kinematic equations?

This is all self study at this point, I am starting college in august, but thought I should go over this stuff before then, as my high school was...less then stellar.
You can't apply the constant acceleration equations here because if you integrate over a full cycle, you'd get a net acceleration of 0. You can't just integrate from 0 to pi/4, use that average for constant acceleration, and say that the equation you have models the displacement, even if you mean to say that it models it in the time frame you supplied. In this case, you're not even working with time frames as you're integrating with respect to x. That brings up even more problems as you're take work, dividing it by the net displacement, calling that the average force function, and then solving for a general acceleration.

Like the rest of the people posting here, I'm confused as to why you're using formulas from other topics here. I really hate sounding so much like a butt hole by pointing out those errors, but they're pretty blatant.

My advice to you is that if you're tryign to self study calculus-based mechanics or E&M, you should check out MIT's Open Courseware Lectures for physics. Walter Lewin, the professor in the lecture videos, is perhaps one of the best lecturers I've ever seen. He thoroughly explains it. You can get the accompanying textbook while you're at it. It would be an old edition, so the cost should be no more than a few bucks if even.
 
What solution "in textbooks" are you referring to?
[tex]x(t)=Acos(\omega_{0}t+\phi)[/tex]

I mean I get where that comes from and how you can let the constants equal Asin(phi) and Acos(phi). I just meant I've never seen the actual step-by-step math to get there.
 

cepheid

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[tex]x(t)=Acos(\omega_{0}t+\phi)[/tex]

I mean I get where that comes from and how you can let the constants equal Asin(phi) and Acos(phi). I just meant I've never seen the actual step-by-step math to get there.
It comes from the sum and difference formulas for trig functions, one case of which states that:

[tex] \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) [/tex]​

In this case, A = [itex] \omega t [/itex] and B = [itex] \phi [/itex]

Expanding this, we obtain:

[tex] A\cos(\omega t + \phi) = A\cos(\omega t)\cos(\phi) - A\sin(\omega t)\sin(\phi) [/tex]​

Now we set [itex] C_1 = A\cos(\phi) [/itex] and [itex]C_2 = -A\sin(\phi) [/itex]

The result is:

[tex] A\cos(\omega t + \phi) = C_1\cos(\omega t) + C_2 \sin(\omega t)[/tex]​

This is a statement of the result that ANY sinusoidal waveform can be expressed as a linear combination of a sine wave and a cosine wave. The constants in that linear combination determine the initial phase ([itex] \phi [/itex]) which in this case tells you the starting position and velocity of the mass on the spring. IF the mass starts out at t = 0 at the equilibrium position, with maximum speed, then this corresponds to a sine wave only (just look at a sine wave at t = 0 to confirm this). In this case, [itex] C_1 = 0 [/itex] i.e. [itex] \phi = - \pi /2 [/itex].

If the mass starts out at one extreme end of the motion at t = 0 (maximum displacement and zero speed), then clearly this corresponds to a cosine wave only (just look at a cosine wave at t = 0 to confirm this). In this case, [itex] C_2 = 0 [/itex] i.e. [itex] \phi = 0 [/itex].

The most general case (for arbitrary initial conditions) is something in between.
 
It comes from the double angle formula, which states that:

[tex] \cos(A + B) = \cos(A)\cos(B) - \sin(A)\sin(B) [/tex]​

In this case, A = [itex] \omega t [/itex] and B = [itex] \phi [/itex]

Expanding this, we obtain:

[tex] A\cos(\omega t + \phi) = A\cos(\omega t)\cos(\phi) - A\sin(\omega t)\sin(\phi) [/tex]​

Now we set [itex] C_1 = A\cos(\phi) [/itex] and [itex]C_2 = -A\sin(\phi) [/itex]

The result is:

[tex] A\cos(\omega t + \phi) = C_1\cos(\omega t) + C_2 \sin(\omega t)[/tex]​

This is a statement of the result that ANY sinusoidal waveform can be expressed as a linear combination of a sine wave and a cosine wave. The constants in that linear combination determine the initial phase ([itex] \phi [/itex]) which in this case tells you the starting position and velocity of the mass on the spring. IF the mass starts out at t = 0 at the equilibrium position, with maximum speed, then this corresponds to a sine wave only (just look at a sine wave at t = 0 to confirm this). In this case, [itex] C_1 = 0 [/itex] i.e. [itex] \phi = - \pi /2 [/itex].

If the mass starts out at one extreme end of the motion at t = 0 (maximum displacement and zero speed), then clearly this corresponds to a cosine wave only (just look at a cosine wave at t = 0 to confirm this). In this case In this case, [itex] C_2 = 0 [/itex] i.e. [itex] \phi = 0 [/itex].

The most general case (for arbitrary initial conditions) is something in between.
Thanks! You answered my question from another thread in the Math section, haha. I never learned that you could express sinusoidal functions as a linear combination of sin and cos. Thanks again.
 

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