Question

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?

Answer #1

a) Solving for maximum speed:

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

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

v = 0.5456 m/sec ANSWER

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

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