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Simple Harmonic Motion of a wave

  1. Apr 5, 2008 #1
    [SOLVED] Simple Harmonic Motion

    1. The problem statement, all variables and given/known data
    A cheerleader waves her pom-pom in simple harmonic motion with an amplitude of 18.0 cm and a frequency of 0.850 Hz. Find (a) the maximum magnitude of the acceleration and of the velocity; (b) the acceleration and speed when the pom-pom's coordinate is x = +9.0 cm; (c) the time required to move from the equilibrium position directly to a point 12.0 cm away. (d) Which of the quantities asked for in parts (a), (b), and (c) can be found using the energy approach, and which ones cannot? Explain.


    2. Relevant equations
    abs(a_max)= ω^2A, A = amplitude
    abs(v_max)=ωA
    ω=2πf, f=frequency
    a = -ω^2 sqrt(A^2 - x^2)



    3. The attempt at a solution
    (a) ω=2π(0.85)= 5.34 rad/s
    abs(a_max)= (5.34)^2(0.18)= 5.13m/s^2
    abs(v_max)= (5.34)(0.18) = 0.961 m/s

    (b) a = -(5.34)^2 sqrt((0.18)^2 - (0.09)^2) = -4.45m/s^2
    speed = abs(v) = (5.34) sqrt((0.18)^2 - (0.09)^2) = 0.832 m/s

    I don't know how to approach (c) and (d).
     
  2. jcsd
  3. Apr 6, 2008 #2
    Maybe this will do for the (c) part:
    [tex]x=As in\omega t[/tex].

    (d) part seems easy. Just try calculating every part using the law of conservation:
    [tex]E=E_{k}+E_{p}=E_{k(max)}=E_{p(max)}[/tex].
     
    Last edited: Apr 6, 2008
  4. Apr 8, 2008 #3
    for d) I'm thinking that part (a) can be found by using the energy approach because K_max = (1/2)m(v_max)^2, from v_max, you can find a_max; part (b) cannot be found by using the energy method because energy doesn't deal with what happens in the beginning and end; part (c) cannot be found using the energy approach because the time variable doesn't occur in the energy equations.

    Is that right?
     
  5. Apr 9, 2008 #4
    The velocity can be found at any given point through conservation of energy.
     
  6. Apr 9, 2008 #5
    I'm still a little confused about the energy approach in part (d).
     
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