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Simple Harmonic Motion

  • Thread starter lando45
  • Start date
  • #1
80
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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.

shm.gif


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!
 

Answers and Replies

  • #2
80
0
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.
 
  • #3
andrevdh
Homework Helper
2,128
116
Sometimes the universe do listen to our problems and have pity on us.
 

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