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Yangfan (Tony) Sun26559147Lab Section: LD2
Physics 101 Learning Object 1Energy in a Mass Spring System
Question: A mass is connected a spring oscillator. The mass is pulled back from the spring at a distance of 40.0 cm, is released, and begins to oscillate. The spring constant has a value of 310 kg/s2.
What is the total amount of energy of this system?
During its oscillation, what will its top speed be? Assume the mass is 2.15 kg.
The mass is 25 cm away from the spring’s stable point. What is the speed of it at this moment?
How far from the stable position will the mass be if its kinetic energy is equal to its potential energy? To the left of equilibrium or to the right of equilibrium?
If the mass undergoes simple harmonic motion with an amplitude of 30 cm and is 10 cm away from the spring’s stable position, how much of the total energy is in the form of potential energy?
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Yangfan (Tony) Sun26559147Lab Section: LD2
Physics 101 Learning Object 1Energy in a Mass Spring System
(Solutions)
Question: A mass is connected a spring oscillator. The mass is pulled back from the spring at a distance of 40.0 cm, is released, and begins to oscillate. The spring constant has a value of 310 kg/s2.
What is the total amount of energy in this system?
If we take any position of the mass its path, we can calculate the total energy (E = K + U) because energy is conserved. The only position clearly stated in the question is at its initial position at t = 0. This means there is no kinetic energy. All of the energy is in the form of potential energy. We know
U = ½ kx2
U = ½ (310) (0.400)2
U = 24.8 J.
The total amount of energy in this system is 24.8 J.
During its oscillation, what will its top speed be? Assume the mass is 2.15 kg.
If the mass is moving at top speed, then all the energy of this system in the form of kinetic energy. Using the formula:
v = √ (2E/m)
v = √ (2(24.8)/(2.15))
v = 4.80 m/s
The mass is 25 cm away from the spring’s stable point. What is the speed of it at this moment?
A simple use of the total energy equation can allow us to solve this problem.
E = U + K
E = ½ kx2 + ½ mv2
v2 = (2E – kx2)/m
v = √ [(2E – kx2)/m]
v = √ [(2(24.8) – (310)(0.25)2)/2.15]
v = 3.75 m/s
The speed at this moment is 3.75 m/s.
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Yangfan (Tony) Sun26559147Lab Section: LD2
(Solutions continued)
How far from the stable position will the mass be if its kinetic energy is equal to its potential energy? To the left of equilibrium or to the right of equilibrium?
Using a graph:
(Picture from “Physics for Scientists and Engineers An Interactive Approach” textbook)
We can see that the graph on the right depicts the kinetic energy and potential energy as a function of displacement (in the form x(t) = A cos(ωt + φ). The graph is an even function, and thus the only time the kinetic energy and potential energy are equal is when it is half the total energy, between –A and the equilibrium and between the equilibrium and A, or 24.8/2 = 12.4 J.
K = E
½ kx2 = E
x = √ (2E/k)
x = √ (2(12.4)/310)
x = 0.283 m
The graph shows that the same amount of kinetic energy is the same for equal distances to both the left and right of the equilibrium. This also applies to potential energy.
If the mass undergoes simple harmonic motion with an amplitude of 30 cm and is 10 cm away from the spring’s stable position, how much of the total energy is in the form of potential energy?
U = ½ kx2
E = ½ kA2
U/E = (½ kx2) / (½ kA2)
U/E = x2 / A2
U/E = 102 / 302
U/E = 1/9
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Yangfan (Tony) Sun26559147Lab Section: LD2So 1/9 or about 11% of the total energy is in the form of potential energy when the mass is 10 cm away from the stable position.
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