Question

Constants A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. Att 0 the block has velocity -4.00 m/s and displacement +0.200 m.
I need to know how to do part c only.

Answer 1

The equation for displacement

x(t) = A*cos(w*t + $\phi$ ) ..............eq1

velocity , v = dx/dt = -A*w*sin(w*t + $\phi$ ) ................eq2

angular frequency , w = sqrt(k/m) = sqrt(300/2)

w = 12.24 rad/s

put x = 0.2 at t = 0 in eq1

0.2 = A*cos(0 + $\phi$ )

A = 0.2/cos $\phi$ .............eq3

put v = -4 m/s , w = 12.24 rad /s and A = 0.2/cos $\phi$ in eq2

-4 = - 0.2/cos $\phi$ * 12.24*sin(12.24*0 + $\phi$ )

tan $\phi$ = 4/2.44

$\phi$ = 58.47^o

$\phi$ = 1.02 rad

from eq 3

A = 0.383 m

so , the equation for the displacement is ,

x(t) = 0.383 *cos(12.24*t + 1.02)