Calculating Maximum Net Force in Simple Harmonic Motion

[Dark Visitor]

The position of a .64 kg mass undergoing simple harmonic motion is given by x(t) = (.16 m)cos(\pit/16). What is the maximum net force on the mass as it oscillates? (hints: First, match the quantities in the equation above to x(t) = Acos(\omegat). Next, recall that the max. net force equals the product of the mass and the maximum acceleration.)

* 3.9 x 10^-3 N
* 9.9 x 10^-3 N
* 1.3 x 10^-3 N
* 6.3 N


I don't really understand this problem and I need some help. So if anyone would be willing to help, I would really appreciate it. Where do I begin?

 

[rock.freak667]

well you know F=ma.

You don't have an expression for a. You know x(t)=0.16cos(\frac{\pi t}{16}). Can you find a(t)?

 

[Dark Visitor]

I don't understand how. Would it look anything like the x(t) one?

 

[rock.freak667]

I don't understand how. Would it look anything like the x(t) one?

x is a displacement. Do you know how to get velocity from displacement?

 

[Dark Visitor]

Not that I know of. Please explain it to me.

 

[rock.freak667]

Not that I know of. Please explain it to me.

velocity is the rate of change of displacement: v= dx/dt

acceleration is the rate of change of velocity. Can you find acceleration now?

 

[Dark Visitor]

So: a = dv/dt?

But how does that help me with this?

 

[denverdoc]

Maybe try working on one of these problems at a time ;-D.

Acceleration and force will be maximal when a is at a maximum. Do you know how to do derivatives and do max/min problems?

 

[Dark Visitor]

No I do not.

 

[rock.freak667]

No I do not.

well find a(t). Now think, cosine and sine are maximum when they equal to what number?

 

[Dark Visitor]

I still don't understand how to do that when we don't know x, t, or v.

 

[rock.freak667]

I still don't understand how to do that when we don't know x, t, or v.

x=0.16cos(\frac{\pi t}{16})


do you understand how to differentiate a function?

 

[Dark Visitor]

No. How do I do that, and how does that help me with this (I mean where do I apply it)?

 

[rock.freak667]

No. How do I do that, and how does that help me with this (I mean where do I apply it)?

The entire problem relies on your knowledge of calculus. I must ask, if you do not know how to find the derivative of a function, how did you get this problem as homework?

 

[Dark Visitor]

I don't know any calculus. I have only taken Algebra and Trig., and my teacher gave me these, so they must be using things I know, or should know. Any way you can explain that more in terms I can understand?

 

[rock.freak667]

I don't know any calculus. I have only taken Algebra and Trig., and my teacher gave me these, so they must be using things I know, or should know. Any way you can explain that more in terms I can understand?

read about is http://www.intmath.com/Differentiation/Differentiation-intro.php"

 

[Dark Visitor]

Okay, I think I understand it a little better now. But where and how do I apply that to this problem?

 

[rock.freak667]

I don't know any calculus. I have only taken Algebra and Trig., and my teacher gave me these, so they must be using things I know, or should know. Any way you can explain that more in terms I can understand?

Okay, I think I understand it a little better now. But where and how do I apply that to this problem?

read http://www.analyzemath.com/calculus/Differentiation/trigonometric.html" as well

 

[Dark Visitor]

Okay, but I am still completely lost on this...

 

[rock.freak667]

Okay, but I am still completely lost on this...

Read up on differentiation some more. But for this problem you need to know these

y=sinkx, dy/dx=kcoskx

y=coskx, dy/dx=-ksinkx.

With these can you find a(t) given x(t)?

 

[Dark Visitor]

Well, reading the problem over, it says to match the quantities in the above equation, which leaves me with:

A = .160 m and \omega = \pi/16

But I still don't understand where your differentiations come into play.

 

[rock.freak667]

Well, reading the problem over, it says to match the quantities in the above equation, which leaves me with:

A = .160 m and \omega = \pi/16

But I still don't understand where your differentiations come into play.

Because if you have x=Acos(ωt) and you want to find F_max=ma_max, chances are you will need 'a' to get 'a_max'

 

[Dark Visitor]

Okay, I see what you're saying. And it makes sense lol. But where do I go from here to get to a or a_max?

 

[rock.freak667]

Okay, I see what you're saying. And it makes sense lol. But where do I go from here to get to a or a_max?

Find 'a' and then you can get 'a_max' from 'a'

 

[Dark Visitor]

Well, do I need to use the equation:

a(t) = -w²Acos(wt) ?

 

[rock.freak667]

Well, do I need to use the equation:

a(t) = -w²Acos(wt) ?

yes that is what you'd get had you differentiated x(t).

so now that you know a(t). At what value,cos(ωt) maximum?

 

[Dark Visitor]

In my notes, I have

a_max = w²a

Is that what you are going for?

 

[rock.freak667]

In my notes, I have

a_max = w²a

Is that what you are going for?

yes, but you did not follow the question template, so I really did not know what equations you had.

Had x been x=t⁵+e^tsin2t, you would need to differentiate.

 

[Dark Visitor]

Sorry about that. I found it buried in my notes, and when I posted this, I didn't think I would need that, so I didn't have my notes out.

So what do I do now?

 

[rock.freak667]

Sorry about that. I found it buried in my notes, and when I posted this, I didn't think I would need that, so I didn't have my notes out.

So what do I do now?

a_max=ω²A.

F_max=ma_max

I think you can get F_max from here

 

[denverdoc]

Dark visitor, do you see the similarity between this problem and the engine problem. There we were looking for max velocity. So when you differentiate position with respect to t, the w pops out in front (chain rule) and sin becomes cos. D/dx again and you get -w*w*sin.

Next time you do this, please have all your notes out and for good heavens, my man, don't try to bite off so much in a weekend. Physics requires slow assimilation and lots of practice!

 

[Dark Visitor]

Only one more thing, sorry.

Is the amplitude (A) .160 m? And is \omega equal to\pi/16?

 

[rock.freak667]

Only one more thing, sorry.

Is the amplitude (A) .160 m? And is \omega equal to\pi/16?

yes.

 

[Dark Visitor]

Okay, thanks. Here is my work:

a_max = (\pi/16)²(.160 m)
= .00617 m/s²

F_max = (.64 kg)(.00617 m/s²)
= .003949

or 3.9 * 10^-3

Thanks a lot. I really appreciate all of your help.