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Simple Harmonic Motion vibration Question

  1. Nov 3, 2011 #1
    1. The problem statement, all variables and given/known data
    A mass of 0.3kg is suspended from a spring of stiffness 200Nm-1. If the mass is displaced by 10mm from its equilibrium position and released, for resulting vibration, calculate:
    a) the frequency of vibration
    b) the maximum velocity of the mass during the vibration
    c) the maximum acceleration of the mass during the vibration



    2. Relevant equations
    Angular frequency= √((Spring constant)/mass)
    Frequency= (Angular frequency)/(2 × π)
    Velocity: v(t) = -ωA sin(ωt + φ)
    Acceleration: a(t) = -ω 2 A cos(ωt + φ)

    Displacement = amplitude x sin(angular frequency x time)
    T=1/f

    3. The attempt at a solution
    a) Angular frequency= √((Spring constant)/mass)
    ω_n= √(k/m)
    ω_n= √(200/0.3)
    ω_(n )=25.8199

    Frequency= (Angular frequency)/(2 × π)
    F= ω_n/2π
    F= 25.8199/2π
    F=4.11Hz

    b) Velocity= -angular frequency ×amplitude ×sin⁡〖(angular frequency ×time + phase constant〗
    v(t)= -ωA sin⁡(ωt+φ)
    angular frequency=25.8199
    amplitude=0.01m
    phase constant= ?
    time = 0.243


    My physics is pretty poor i'm afraid, and i'd be grateful for any help possible. I've become stuck trying to calculate the phase constant. I've read that through using x = A*sin(ωt + φ), and setting t to 0, I can calculate what the phase constant should be. I seem to keep getting 0 as my result here however.Any help on how the phase constant should be calculated would be grately appreciated. Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 3, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The "phase constant" is pretty much whatever you want to make it. Specifically, it depends upon what you want to assign as t= 0. However, you first assert that the velocity is given by [itex]-\omega sin(\omega t+ \phi)[/itex] but then you want [itex]x= A sin(\omega t+ \phi)[/itex]. Those can't both be true! The velocity is the derivative of position so if the velocity is given by a sine then x must be a cosine. [itex]x(t)= A cos(\omega t+ \phi)[/itex].

    If you choose to make t= 0 at the beginning of the motion, then you must have [itex]Acos(\omega(0)+ \phi)= Acos(\phi)= A[/itex], the maximum motion. Yes, that reduces to [itex]cos(\phi)= 1[/itex] so [itex]\phi= 0[/itex]. What's wrong with that?
     
  4. Nov 3, 2011 #3
    Thanks for your reply. So I have:

    v(t)= -ωA sin⁡(ωt+φ)
    angular frequency=25.8199
    amplitude=0.01m
    time= 1/f=0.2433
    phase constant= 0
    v(t)= -25.8199 ×0.01 sin⁡(25.8199 × 0.2433+0)
    v(t)= -0.258199 sin⁡(6.2817627+0)
    v(t)= -0.02825m/s

    Is there now further steps involved in calculating the maximum velocity?
     
  5. Nov 3, 2011 #4
    Any body able to offer any more help please.

    Where i'm up to:

    v(t)= -ωA sin⁡(ωt+φ)
    angular frequency=25.8199
    amplitude=0.01m
    time= 1/f=0.2433
    phase constant= 0
    v(t)= -25.8199 ×0.01 sin⁡(25.8199 × 0.2433+0)
    v(t)= -0.258199 sin⁡(6.2817627+0)
    v(t)= -0.02825ms^(-1)

    Now i seem to have confused myself and could really do with some help to point me in the right direction.

    I need to calculate what the maximum velocity would be. I've read that Maximum and minimum values of any sine and cosine function are +1 and -1. So, would the correct equation be v(t)= -0.258199 sin⁡(1)? Furthermore I've read velocity is at a maximum when displacement is zero. By my understanding this would be when the amplitude is at 0.00. In which case should the equation be v(t)= -25.8199 ×0.00 sin⁡(25.8199 × 0.2433+0).
    Both ways seem to give a velocity of 0.00ms?
    And is there any other factors that then need adjusting if amplitude is changed? doesn't seem to be, but thought i'd check.

    If anybody is able to offer their help that would be great. Thanks.
     
  6. Jan 2, 2012 #5
    Rikiki,

    Maximum Velocity is given using

    vmax = Aωcos(ωt+∅)

    Maximum Acceleration is given using

    amax = Aω2sin(ωt+∅)

    Other area of this question which I'm struggling with is " (iv) the mass required to produce double the maximum velocity calculated, using the same spring and initial deflection.

    Hope this helps a little
     
  7. Jan 2, 2012 #6
    The question asks for max velocity and max acceleration.
    Without worrying about phase angle the max velocity occurs as the object passes through the equilibrium position and is given by
    v = ωA (A = amplitude)
    SWimilarly max acceleration occurs at the max displacement
    a = ω^2. A
     
  8. Jan 9, 2012 #7
    im having trouble with;

    Plot a graph of acceleration against displacement (x) (for values of x
    from x = –10 mm to x = +10 mm)
     
  9. Jan 13, 2012 #8
    So far I've worked out
    f=4.11Hz
    vmax=0.26ms-1
    amax=0.73ms-2
    mass to cause double velocity=0.0755kg

    maybe someone can verify these answers.

    I'm stumped on this graph though
     
  10. Jan 13, 2012 #9
    ok forget that, those answers don't work out
    f=4.11hz
    vmax=0.257ms-1 using either equation above
    amax=6.67ms-2 again using either equation

    for the graph we can use a= -(2 pi f)*2 xdisplacement to find acceleration at any point during the oscillation.
     
  11. Jan 13, 2012 #10
    In #8 you have max acceleration = 0.73 and in fact it is 6.67 as stated in #9
    in SHM acc is proportional to displacement so the graph is a straight line passing through the origin and a = 6.67 when A(amplitude) = +/-10mm
     
    Last edited: Jan 13, 2012
  12. Jan 13, 2012 #11
    thanks technician i got that eventually. for the graph, the value is 6.67 at -10mm straight line passing through zero to -6.67 at +10mm using a=-(2pif)^2 x to find your other acceleration values at displacement points during oscillation
     
  13. Apr 17, 2012 #12
    Hi

    Can anyone please explain how the graph is drawn.

    Thanks in advance
     
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