Question

A 0.31-kg object connected to a light spring with a force constant of 24.6 N/m oscillates on a frictionless horizontal surface. The spring is compressed 4.0 cm and released from rest.

(a) Determine the maximum speed of the object.
m/s

(b) Determine the speed of the object when the spring is compressed 1.5 cm.
m/s

(c) Determine the speed of the object as it passes the point 1.5 cm from the equilibrium position.
m/s

(d) For what value of x does the speed equal one-half the maximum speed?
m

Answer 1

A) maximum speed is Vmax = A*w =

here A is the amplitude = 4 cm =0.04 m

angular speed is w = sqrt(k/m) = sqrt(24.6/0.31) = 8.908 rad/s

Vmax = 0.04*8.908 = 0.356 m/s

B) From law of conservation of energy
Total energy at mean position = Total energy at x= 1.5 cm

0.5*K*A^2+0 = 0.5*m*v^2+0.5*k*x^2

0.5*24.6*0.04^2 = 0.5*0.31*v^2+(0.5*24.6*0.015^2)

0.5*0.31**v^2 = 0.0169125

v = sqrt(0.0169125)/(0.5*0.31) = 0.33 m/s

C) v = w*sqrt(A^2-x^2)

v = 8.908*sqrt(0.04^2-0.015^2)
v = 0.33 m/s

D) From law of conservation of energy
Total energy at mean position = Total energy at x= 1.5 cm

0.5*K*A^2+0 = 0.5*m*v^2+0.5*k*x^2

0.5*24.6*0.04^2 = 0.5*0.31*v^2+(0.5*24.6*x^2)

here given v = Vmax/2 = 0.356/2= 0.178
0.01968 = 0.00491102+(12.3*x^2)

12.3*x^2 = 0.01476898

x = sqrt(0.01476898/12.3) = 0.0346 m = 3.46 cm