Answer to Question #231867 in Physics for Uncle Rex

Question

For a system of mass m=0.05kg and k=50Nm-¹ at time t=0, the mass is observed to be displaced 60mm to the right of its equilibrium position and to be moving to the right at the speed of 2.5ms-¹. Calculate:

a)the amplitude

b)the phase constant

c)the energy

Answer 1

a)

$$kA^2=kx^2+mv^2\\50A^2=50(0.06)^2+(0.05)(2.5)^2\\A=0.099\ m$$

b)

$$0.06=0.099\cos {\phi}\\\phi= 0.92\ rad$$

c)

$$E=0.5(50)(0.099)^2=0.245\ J$$