SHM: Which is amplitude and which is displacement?

  • Thread starter richyr33
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  • #1
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Homework Statement


Calculate acceleration and velocity of a piston. Stroke is 150mm and engine speed 3500rpm
Using simple harmonic motion equations

Homework Equations


a = -ω^2*s
v = ω*√A^2 - s^2
ω = 2*∏*f

v = velocity, a = acceleration, f = frequency
ω=angular velocity, s=displacement, A = Amplitude


The Attempt at a Solution


worked out frequency by 3500/60=58.3Hz
angular velocity = 366.31 radians/sec

Problem i have is i'm not sure what value i use for amplitude and what value i use for displacement. I believe that amplitude would be half the stroke thus being 75mm and displacement being the stroke of 150mm. However inserting these values into the equation to calculate velocity gives a negative figure which then cannot be square rooted. Any help would be appreciated. Thanks
 

Answers and Replies

  • #2
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what equations are you using specifically?

Acos(kr-wt)=x? = Acos(wt)=x where r=0 for the wave front.

w= 3500x60 rad/s

when it says stroke does it mean from the max to min or to the middle?

v = -wAsin(wt)

a= -w2cos(wt)


did you write out the full question because it doesn't seem complete imo
 
  • #3
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That is the question in full. It just says the stroke of the engine is 150mm. The only pieces of information I am missing are A=Amplitude and s = displacement. Only thing that is confusing me is the difference between the two. Is amplitude the distance from bottom of stroke to top or is that the displacement?
 
  • #4
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displacement is the distance traveled from the center as every book i've ever seen states it.
 
  • #5
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I think in this case the displacement and amplitude are the same value as the question asks for maximum velocity which would be when displacement is at a maximum, which would be distance from the centre like you say, which is 150mm/2.

I came up with an answer of 27.47m/s for velocity and 10063.73m/s/s for acceleration.

Thanks for your help, appreciated.
 
  • #6
1,506
18
Amplitude is the maximum value of the displacement. I would say that in your example the stroke (150mm) is the amplitude
3500rpm means the frequency is 58.3Hz which means that ω = 2πf = 366....just as you found.
I would take 150mm (0.155m) to be the Amplitude which means that
max acceleration = ω^2*A = ω^2*0.155 = 20762 m/s^2
Max velocity = ωA = 366*0.15 = 55m/s
Hope this helps
 
  • #7
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I think you have to take 'stroke' to mean amplitude, it makes sense to refer all distances from the equilibrium position.
I am certain this is the convention.
 
  • #8
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Yes this is exactly whats confusing me and I have two differing explanations in my notes which adds to it! I have worked it out for the stroke being the amplitude as well and i have the same answers as you have worked out and I have also done it as above where i have taken the amplitude to be the stroke halved.
wouldn't the equilibrium point be half way up the cylinder thus making the stroke twice the amplitude?
 
  • #9
1,506
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RichYr33
What you say is a very good point and could be correct.... I don't know!
Do you know the answer?
Either way you know how to do the calculation which is the important thing
 
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