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Simple Harmonic Motion

  1. Sep 26, 2006 #1
    My teacher set us a bunch of pendulum/SHM questions, and I managed to crack them all except for this one which has been bothering me:

    "A 5.4 g bullet is fired horizontally into a 0.50 kg block of wood resting on a frictionless table. The block, which is attached to a horizontal spring, retains the bullet and moves forward, compressing the spring. The block-spring system goes into SHM with a frequency of 9.5 Hz and an amplitude of 15 cm. Determine the initial speed of the bullet."

    I found the following diagram to help me calculate my answer but I came up with the wrong answer.


    Here's what I did:

    Displacement (y) = Amplitude (A) x sin (Angular Frequency [w] x Time [t])
    y = 0.15 x sin ([2 x Π x f] x [1 / f])
    y = 0.15 x sin (59.7 x 0.105)
    y = 0.15 x 0.109
    y = 0.0164
    So in time 0.105 seconds the block w/ spring moved 0.0164 metres which is equivalent to 0.156ms^-1.

    I then plugged this value into the motion equation m1v2 = m2v2 and got the following:

    0.0054u = 0.5054v
    v = 0.156
    0.0054u = 0.5054 x 0.156
    0.0054u = 0.079
    u = 14.6ms^-1

    I tried submitting this answer but it was wrong...can someone please help me out!
  2. jcsd
  3. Sep 26, 2006 #2
    EDIT: OK I just solved it, I realised I was using the wrong equation, I shoulda been using the differential of the y displacement function. Thanks anyway.
  4. Sep 26, 2006 #3


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