Max Speed of Piston in Simple Harmonic Motion

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The discussion revolves around calculating the maximum speed of a piston in simple harmonic motion, specifically for a 1 kg piston moving 10 cm at 3000 rpm. The participants clarify that the amplitude (A) is 5 cm, derived from the total distance of 10 cm. They confirm the formula for maximum velocity as vmax = ωA, where ω is the angular velocity converted to 314.159 rad/sec. Using these values, the maximum speed is calculated to be 15.71 m/s. The conversation emphasizes understanding the relationship between amplitude, angular velocity, and maximum speed in harmonic motion.
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The motion of a piston in an automobile engine is nearly simple harmonic. If the 1 kg piston travels back and forth over a total distance of 10 cm, what is the maximum speed (magnitude of velocity) when the engine is running at 3000 rpm? (Hint: the 10 cm distance has something to do with amplitude; 3000 rpm needs to be converted to rad/sec)


I did convert the 3000 rpm, and I got 314.159 rad/sec.


I need some help with this problem a.s.a.p. I don't know where to go with it. Thanks for any help you can give me.
 
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What formula do you think you will need here?
 
I don't know, and I don't know any way to find A. I keep thinking that A has something to do with the 3000 rpm, but I am not sure.
 
What formula(s) do you know that would allow you to find the maximum velocity?
 
Dark Visitor said:
I don't know, and I don't know any way to find A. I keep thinking that A has something to do with the 3000 rpm, but I am not sure.

A is the amplitude. If the piston moves, from minimum to maximum, a distance of 10cm, what distance does it move when it goes from 0 to maximum?
 
5 cm? Cause 10 cm would be the total length, while +A is half, and -A is the other half.
 
Dark Visitor said:
5 cm? Cause 10 cm would be the total length, while +A is half, and -A is the other half.

Right good. Now can you formulate an equation for speed given kinetic energy is 1/2mv2 and potential energy is 1/2kx2. What is the formula for the total energy?
 
E = K + U

(K is kinetic energy, u is potential energy)

But using the equations you gave, I am missing most of the quantities. How can I use them?
 
Dark Visitor said:
E = K + U

(K is kinetic energy, u is potential energy)

But using the equations you gave, I am missing most of the quantities. How can I use them?

I told you K and U, if you get an expression for E, you will get what you need with the quantities you know.
 
  • #10
But for K, I need a velocity, which I don't have.

And for U, I need a spring constant value for k, but we have x (10 cm).
 
  • #11
Dark Visitor said:
But for K, I need a velocity, which I don't have.

And for U, I need a spring constant value for k, but we have x (10 cm).

you don't need to plug in all the values at one time. Just formulate and then see if you can get the terms needed.

what is E in terms of k and A?
 
  • #12
Well since I don't have all the values I need, I get:

K = 1/2(1 kg)(v)2

and

U = 1/2(k)(.01 m)2

and then we still have E = K + U
 
  • #13
Dark Visitor said:
Well since I don't have all the values I need, I get:

K = 1/2(1 kg)(v)2

and

U = 1/2(k)(.01 m)2

and then we still have E = K + U

do not put in the values as yet.

E=\frac{1}{2}mv^2+\frac{1}{2}kx^2

Do you know another way to re-write E in terms of k and A?
 
  • #14
I think what is being asked for is an equation that resembles the other oscillators.


x(t)=A Sin(argument in t) where your argument has you hitting both extremes at a rate of 50 times a second, say by multiplying t by something like 100 * pi (314.15)
Now remember to get the max velocity, you need to know when acceleration is zero.
The velocity is -wA sin(wt). sin(wt) is at most 1.0. Hope this helps, I've been at this most of the day and am burnt to a crisp.
 
  • #15
So do I use the equation:

vmax = wA ? Or what? I am really confused on this one...
 
  • #16
Dark Visitor said:
So do I use the equation:

vmax = wA ? Or what? I am really confused on this one...
YES! Vmax=wA. Good job. Sorry if I'm a little fried and testy.
 
  • #17
Dark Visitor said:
So do I use the equation:

vmax = wA ? Or what? I am really confused on this one...


yes, had you derived it, you would get v_{max}=\omega A
 
  • #18
No, don't be. I owe you big time for all of your help today, and I would be the same way if I was helping someone like me.

Now that I got that right, where do I get my w and A values?
 
  • #19
Dark Visitor said:
No, don't be. I owe you big time for all of your help today, and I would be the same way if I was helping someone like me.

Now that I got that right, where do I get my w and A values?

A is the amplitude that you found, and w is the angular velocity you converted in the first post.
 
  • #20
Okay, and I got:

vmax = (314.159 rad/s)(.05 m)
= 15.71 m/s


Thanks a lot, both of you, for your help. I appreciate it more than you know. :redface:
 
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