Total energy of a mass hanging on a spring

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SUMMARY

The discussion centers on calculating the total energy of a mass hanging on a spring undergoing Simple Harmonic Motion (SHM). The user initially calculated potential energy (E_p) and kinetic energy (E_k) as both equaling 0.125 J, leading to a total energy (E_t) of 0.25 J. However, the correct total energy is 0.125 J, as clarified by other participants who explained that maximum potential and kinetic energies do not occur simultaneously. The correct velocity formula for SHM is also provided, emphasizing the importance of understanding energy conservation in oscillatory systems.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with potential energy and kinetic energy equations
  • Knowledge of angular frequency in oscillatory systems
  • Ability to interpret energy conservation principles in physics
NEXT STEPS
  • Study the derivation and application of the SHM velocity formula: $$v(x) = \omega(A^2 - x^2)^{1/2}$$
  • Learn about energy conservation in oscillatory motion, focusing on maximum potential and kinetic energy states
  • Explore the relationship between angular frequency and spring constant in SHM
  • Review graphical representations of energy in SHM to visualize potential and kinetic energy changes
USEFUL FOR

Students and educators in physics, particularly those focusing on mechanics and oscillatory motion, as well as anyone looking to deepen their understanding of energy conservation in systems undergoing Simple Harmonic Motion.

arhzz
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Homework Statement
An object with an mass of m = 1 kg hangs on a vertical spring (k = 100 N / m) and vibrates with an amplitude of 0.05 m.

a) What is the maximum potential energy of this oscillator and how large is the total energy?
Relevant Equations
Et = Ep +Ek
Hello!

So here what I did is first calculated the potential energy; $$ E_p = \frac{1}{2} * k * x^2 $$ E_p should be = 0,125 J Now i tried calculating the kinetic energy, I used this formula $$ E_k = \frac{mv^2}{2} $$ to get v I used this formula $$v = x *\sqrt{\frac{k}{m}} $$ v should be = 0,5 m/s. If we plug that back in the formula for kinetic energy get 0,125 J as well. Now the total energy should be the sum of these two which means ## E_t = 0,25 J ## But this was graded wrong by my teacher,the right answer is 0,125 J. But I don't understand how? I'd assume we have no kinetic energy but since the it is vibrating (moving) kinetic energy should exist. What am I missing here?

Thanks!
 
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arhzz said:
$$v = x *\sqrt{\frac{k}{m}} $$
I think this is not correct.
for a a particle performing SHM $$v(x) = \omega(A^2 - x^2)^{1/2}$$
where ##\omega = (k/m)^{1/2}##(the agular frequency) and ##A## is the maximum displacement of the body from its mean position.
 
Hamiltonian299792458 said:
I think this is not correct.
for a a particle performing SHM $$v(x) = \omega(A^2 - x^2)^{1/2}$$
where ##\omega = (k/m)^{1/2}##(the agular frequency) and ##A## is the maximum displacement of the body from its mean position.
Hmm okay that would actually make sence,that my calculation of Ek was wrong.Although I have never seen this formula before,I'll give it a shot. And could you please tell me what SHM means?
 
arhzz said:
a) What is the maximum potential energy of this oscillator and how large is the total energy?
Relevant Equations:: Et = Ep +Ek

I'd assume we have no kinetic energy but since the it is vibrating (moving) kinetic energy should exist. What am I missing here?
If you want the use the maximum potential energy and the maximum kinetic energy to calculate the total energy, you should make sure the maxima are reached at the same time.
 
You have worked out the maximum ke and the maximum ke. However, you can't simply add them to find the total energy, because when the ke is max, the pe is minimum and vice versa.

Max v and Max ke at equlibrium position
Max pe at max dispacement, where v and hence ke are instantaneously zero.
 
willem2 said:
If you want the use the maximum potential energy and the maximum kinetic energy to calculate the total energy, you should make sure the maxima are reached at the same time.
You don't really mean this, do you? How can the maxima be reached at the same time?
 
  • #10
kuruman said:
You don't really mean this, do you? How can the maxima be reached at the same time?
I said, that the maxima had to occur at the same time IF you wanted to add the maximum potential and kinetic energy
 
  • #11
willem2 said:
I said, that the maxima had to occur at the same time IF you wanted to add the maximum potential and kinetic energy
That's not what you said in post #6. In any case, the maxima never occur at the same time so why even bother mentioning the addition of the two maxima? I am trying to see what your point is.
 
  • #12
kuruman said:
That's not what you said in post #6. In any case, the maxima never occur at the same time so why even bother mentioning the addition of the two maxima? I am trying to see what your point is.
Yes, that is precisely what I said in post #6, and also what appears to have gone wrong in #1.
 
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  • #13
OK, I see your point. Thanks.
 
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