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SHM: Which is amplitude and which is displacement?

  1. Feb 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate acceleration and velocity of a piston. Stroke is 150mm and engine speed 3500rpm
    Using simple harmonic motion equations

    2. Relevant 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

    3. 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
  2. jcsd
  3. Feb 16, 2012 #2
    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
  4. Feb 16, 2012 #3
    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?
  5. Feb 16, 2012 #4
    displacement is the distance traveled from the center as every book i've ever seen states it.
  6. Feb 16, 2012 #5
    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.
  7. Feb 16, 2012 #6
    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
  8. Feb 16, 2012 #7
    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.
  9. Feb 16, 2012 #8
    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?
  10. Feb 17, 2012 #9
    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
    Last edited: Feb 17, 2012
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