Question

# A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in...

A 0.400-kg object attached to a spring with a force constant of 8.00 N/m vibrates in simple harmonic motion with an amplitude of 12.2 cm.

the maximum value of its speed is 54.6
WHAT IS THE MAXIMUM VALUE OF IT'S ACCELERATION?

QUESTION 2
A 45.0-g object connected to a spring with a force constant of 40.0 N/m oscillates with an amplitude of 7.00 cm on a frictionless, horizontal surface.

the total energy of the system is 98
the speed of the object when its position is 1.30 cm is 2.05
(c) Find the kinetic energy when its position is 3.50 cm.
(d) Find the potential energy when its position is 3.50 cm.

QUESTION 3
A 1.50-kg object attached to a spring moves without friction (b = 0) and is driven by an external force given by the expression F = 5.60sin(2?t), where F is in newtons and t is in seconds. The force constant of the spring is 29.0 N/m.

resonance angular frequency of the system is 4.4
the angular frequency of the driven system is 6.28
WHAT IS THE AMPLITUDE OF MOTION?

a) Solving for maximum speed:

v = ?[{(Xo)^2 - (X^2)}k/m]
v = ?[{(0.122)^2 - (0)^2}(8/0.40)]

Solving for maximum acceleration:
a = - (k/m)Xo
a = - (8/0.40)(0.122)
a = - 2.44 m/sec^2

b) For this system, since the amplitude is 7.00 cm, the maximum compression of the spring is 7.00 cm. Let's assume this oscillation started by compressing the spring 7.00 cm and releasing the object attached to it (no initial velocity).

(a)
E = U(initial) = kx^2/2
E = (40.0 N/m)(0.0700 m)^2 / 2
E = 0.098 J = 98 mJ

(b)
E = U + K
E = kx^2/2 + mv^2/2
0.098 J = (40.0 N/m)(0.013 m)^2 / 2 + (0.045 kg)(v^2) / 2
v = 2.05 m/s

(c)
E = kx^2/2 + K
0.098 J = (40.0 N/m)(0.0350 m)^2 / 2 + K
K = 0.0735 J = 73.5 mJ

(d)
U = kx^2/2
U = (40.0 N/m)(0.0350 m)^2 / 2
U = 0.0245 J = 24.5 mJ