A 5 kg object is hanging from an ideal massless spring attached to the ceiling.

Question

A 5 kg object is hanging from an ideal massless spring attached to the ceiling. The object undergoes vertical simple harmonic motion after being displaced by 10 cm. Take the elastic potential energy to be zero for the unstretched spring and the gravitational potential energy to be zero at the lowest point of the motion. At the highest point of the motion, the spring has its natural unstretched length. find the elastic potential energy, the gravitational potential energy and the kinetic energy at the following points.

A) top of the motion

B) the equilibrium position

C) a point 2 cm above the bottom of the motion

D) Find the period of oscillation.

Explanation / Answer

Initial  elastic potential energy, = 1/2 * kx^2

(a)
At the highest point of the motion, the spring has its natural unstretched length.

So,
Elastic potential energy = 0
Kinetic Energy = 0

Gravitational Potential Energy = m*g*h
Gravitational Potential Energy = 5 * 9.8 * 0.1 J
Gravitational Potential Energy = 4.9 J

(b)
Equilibrium Position,
Using Energy Conservation,
1/2 * kx^2 = 4.9
1/2 * k * 0.1^2 = 4.9
k = 980 N/m

Now,
F = K*x
m*g = K*x
5.0 * 9.8 = 980 * x
x = 0.05 m
This is the equilibrium position,

Now,
Gravitational Potential Energy = m*g*h
Gravitational Potential Energy = 5.0 * 9.8 *(0.1-0.05) J
Gravitational Potential Energy = 2.45 J

Elastic potential energy = 1/2 * kx^2
Elastic potential energy = 1/2 * 980 * 0.05^2 J
Elastic potential energy = 1.225 J

Kinetic Energy = 4.9 - 2.45 - 1.225 J
Kinetic Energy = 1.225 J

(c)
Gravitational Potential Energy = m*g*h
Gravitational Potential Energy = 5.0 * 9.8 *(0.02) J
Gravitational Potential Energy = 0.98 J

Elastic potential energy = 1/2 * kx^2
Elastic potential energy = 1/2 * 980 * (0.1 - 0.02)^2 J
Elastic potential energy = 3.136 J

Kinetic Energy = 4.9 - 3.136 - 0.98 J
Kinetic Energy = 0.784 J

(d)
T = 2*3.14 * sqrt(m/k)
T = 6.28 * sqrt(5.0/980)
T = 0.448 s

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