A 1.2-kg block rests on a frictionless surface and is attached to a horizontal s

Question

A 1.2-kg block rests on a frictionless surface and is attached to a horizontal spring of constant k = 40 N/m (see the figure) (Figure 1) . The block oscillates with amplitude 10 cm and phase constant =/2. A block of mass 0.80 kg moves from the right at 1.7 m/s and strikes the first block when the latter is at the rightmost point in its oscillation. The two blocks stick together. I found the frequency to be .71Hz and the amplitude to be 18cm. Im not sure if that helps find phaze. Im just so confused on the phaze constant stuff Determine the phase constant (relative to the original t=0) of the resulting motion.

Explanation / Answer

SHM equation
x(t) = A*sin(w*t + phi)

a) w = sqrt(k/m)

m = m1 + m2 = 1.2 kg + 0.8 kg = 2.0 kg

w = sqrt(40 N/m / 2.0 kg) = 4.47 rad/s

b) This part use energy balance
Use the energy of both components to find the maximum displacement
Max spring PE for the 1.2 kg block + KE for the 0.8 kg block = the new maximum spring PE
1/2*k*A1^2 + 1/2*m2*v^2 = 1/2*k*A2^2
A2 = sqrt(2/k*(1/2*k*A1^2 + 1/2*m2*v^2))
A2 = sqrt(A1^2 + m2/k*v^2)

I got  
A2 = 0.1300 m or 13 cm

c) Use x(t) equation above with the new A, w, and initial displacement to find phi
x(0) = 10 = 13 * sin(3*0 + phi)
phi = asin(10/13)
phi = 0.878 rad

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