# Simple Harmonic Motion

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