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Simple Harmonic Motion & Pendulum Problems! Need HELP!

  1. Feb 14, 2008 #1
    Question 1:
    A 0.1 kg block attached to a spring of force constant 16.9 N/m oscillates with an amplitude of 6 cm. Find the maximum speed of the block. Answer in units of m/s.

    Question 2:
    Find the speed of the block when it is 3 cm from the equilibrium position. Answer in units of m/s.

    Question 3:
    Find its acceleration at 3 cm from the equilibrium position.
    Answer in units of m/s2.


    1) Vmax = .06(16.9/.1)^.5
    = 0.78 THIS IS CORRECT

    2) v = Vmaxsin(pi/3)
    = 0.78(pi/3)
    = THIS IS WORONG

    3) a = .5Vmaxomega
    = .5(7.8) (16.9/.1)^.5
    = THIS IS WRONG TOO

    I tried and I got 2 and 3 wrong :( I ran out of solutions. help please

    I was pretty sure one was right.. but it said it was wrong when I entered it :(
     
    Last edited: Feb 14, 2008
  2. jcsd
  3. Feb 14, 2008 #2

    rock.freak667

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    Homework Helper

    Do you know a formula to find the maximum speed if x=Asin[itex]\omega t[/itex] ?
     
  4. Feb 14, 2008 #3
    Re:

    Are you saying that is the speed formula? I am not sure of any.. I feel completely lost!
     
  5. Feb 15, 2008 #4

    rock.freak667

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    Homework Helper

    The first one is correct.

    For the 2nd one, if [itex]x=Asin\omega t[/itex]
    [itex]\frac{dx}{dt}=v=A\omega cos\omega t[/itex].....(*)

    [tex]sin^2 \omega t+ cos^2 \omega t =1 \Rightarrow cos^2 \omega t=\pm \sqrt{1-sin^2 \omega t}[/tex]

    Making (*) become
    [tex]v=\pm \omega A\sqrt{1-sin^2 \omega t}[/tex]

    [tex]v=\pm \omega \sqrt{A^2-A^2sin^2 \omega t}[/tex]

    [tex]v=\pm \sqrt{A^2-x^2}[/tex]

    For the third part. What equation do you know connecting acceleration and displacement??
     
  6. Feb 15, 2008 #5
    Ok here goes.

    The formula for SHM is x = Asin[wt] (w representing omega/angular momentum)
    w = sqrt(k/m) Just does :P

    So

    x = Asin[sqrt(k/m).t]

    v=x/t or dx/dt
    d(Asin[sqrt(k/m).t])/dt just diffentiates x which is Asin[sqrt(k/m).t], with respect to t
    v = dx/dt = sqrt(k/m).Asin[sqrt(k/m).t]

    The sin function vaires between 1 and -1,
    so v is at a max when sin[sqrt(k/m).t]=1
    so you get
    vMAX = sqrt(k/m).A

    PART 2

    For this part you have the same two equations

    x = Asin[sqrt(k/m).t]
    v = sqrt(k/m).Asin[sqrt(k/m).t]

    You are asked to find out what v is at 3cm.
    However your v equation depends on t so the first
    thing you must do is find at what time t x=3

    x = Asin[sqrt(k/m).t]
    x/A = sin[sqrt(k/m).t]
    sin-1[x/a] = sqrt(k/m).t sin-1 indicates inverse sin
    t = (sin-1[x/a])/sqrt(k/m)

    with this t you just plug it into your eq for v:
    v = sqrt(k/m).Acos[sqrt(k/m).t].

    PART 3

    a = dv/dt so you just diffrentiate your eq for v again
    a = -(k/m).Asin[sqrt(k/m).t]

    Now you know what time x=3 so you just put this in the equation for a

    If you write these down it will be very clear.
     
  7. Feb 16, 2009 #6
    i need help with thia problem i seem to be stuck on finding the spring constant here is the problem
    a spring is hanging with a 50kg mass on its end. when 20g is added it stretches 7cm more. a)find the spring constant b) if the 20g is removed, what i the new period of motion
     
  8. Feb 24, 2009 #7
    a)K=F/x
    k=(.02x9.81)/.07
    k=2.8Nm^-1
    b)f=(1/2[tex]\pi[/tex])([tex]\sqrt{}(k/m)[/tex]
    f=(1/2[tex]\pi[/tex])([tex]\sqrt{}(2.8/50)[/tex]
    f=0.03766Hz
    f=1/t
    t=26.6s

    Does that help? you should create ur own thread 4 different questions...
     
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