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    Part of a building structure is oscillating with approximately simple harmonic motion with a period of 10 s and amplitude 1 m.
    a) calculate the maximum speed of the oscillation.
    b) calculate the maximum acceleration of the oscillation.
    Can you explain how to work these out please as I have no clue, the equations are confusing me.
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    Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there.


    Just quoting in Fox Corner so she can move the thread if needed :wizard:
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    :wizard: popped this in the physics forum for you
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    The relevant equations appear in the formula book as

    vmax=2πfA
    amax=(2πf)2 A

    A (capital) is the amplitude
    f is the frequency of oscillation... which is 1/T (T is the period in seconds)
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    (Original post by amy1999)
    Part of a building structure is oscillating with approximately simple harmonic motion with a period of 10 s and amplitude 1 m.
    a) calculate the maximum speed of the oscillation.
    b) calculate the maximum acceleration of the oscillation.
    Can you explain how to work these out please as I have no clue, the equations are confusing me.
    In simple harmonic motion, the acceleration is always directly proportional to displacement and has the following defining equation:

     a = - \omega^2 x ----eqn(1)

    where  a is the acceleration,
     x is the displacement and
     \omega is the angular frequency , given by  \omega = 2 \pi f = \frac{2 \pi}{T} .

    From eqn(1), the maximum value of acceleration occurs at when the displacement is maximum - maximum displacement implies amplitude, A.

    So the maximum acceleration is

     a = \omega^2 A = (2 \pi f)^2 A ----given by Joinedup

    The displacement of simple harmonic motion can be described by a sinusoidal function

     x = A \cos( \omega t) ----eqn(2)

    If equation (2) is differentiated with respect to time t, we obtain the "velocity"

     v = -\omega A\sin( \omega t) ----eqn(3)

    The maximum speed would occur when  \sin( \omega t) = \pm 1 , so we have

     v = \omega A = (2 \pi f)A ----given by Joinedup
 
 
 
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