I have read that in case of mass spring system max. energy is given by $\frac{1}{2}kX_o^2$. According to this formula energy depends on spring constant and amplitude. But are there any other factors on which it depends? Specifically frequency? I asked someone who said it does but I don't get it how? It is not even mentioned in the formula. And in case of pendulum, what are the factors on which total energy depends? The simple formula is $mgh$ but somewhere I read the formula $mgL (1-\cos\theta)$. Does length of pendulum really play a role here or is it just $\theta$ that determines the value of total energy? Please explain with formulas if possible.
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$\begingroup$ Spring constant $k$ can be written as $m \omega^{2}$ Where $\omega$ is the frequency of oscillation. $\endgroup$– physics101Commented Oct 12, 2017 at 19:15
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2 Answers
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For a mass on a spring SHM, you can write the total energy as: $$E_{tot} = \frac{1}{2}kX_o^2=\frac{1}{2}m \omega^2 X_o^2$$ Because, $\omega=\sqrt\frac{k}{m}$ which rearranges to $k=m\omega^2$, as physics101 commented.
So the total energy depends on the spring constant, the mass, the frequency, and the amplitude. But you don't see them all in the formula at the same time because they are dependent on one another. Specifically, you can determine the spring constant from the mass and frequency.
For the case of a simple pendulum, I will write the total energy as $$E_{tot} = mgL(1-\cos{\theta_o})$$ Where $\theta_o$ is the amplitude. This formula comes from considering the energy of the pendulum when $\theta=\theta_o$, when all of the energy is gravitational potential energy. You could also write $E_{tot}=mgh$, where $h=L(1-\cos{\theta_o})$. Again, to get this in terms of frequency, look at the equation for the frequency of this type of simple harmonic oscillator. For a simple pendulum in the small-angle approximation: $$\omega=\sqrt\frac{g}{L}$$ Where $g$ is the acceleration due to gravity, and $L$ is the length of the string. Rearranging this gives $$g=L\omega^2$$ and so: $$E_{tot}=mL^2\omega^2(1-\cos{\theta_o})$$
In both of these cases, the total energy depends on frequency, but its role is somewhat "hidden" by the typical way we write the formulae.
For your second question, yes the length of the string does play a role in the energy of a pendulum. Compare two pendulums with equal masses and amplitudes ($\theta_o$), but with different length strings. The mass of the pendulum with the longer string will move up farther against gravity, granting it a larger maximum gravitational potential energy when it comes to rest at $\theta=\theta_o$. By conservation of energy, its total energy will be equal to this number, and will therefore also be larger than the pendulum with a short string.
Again, you could change your equation for total energy to eliminate $L$ (just as we removed the dependence on $k$ and $g$ before) but that doesn't mean those parameters don't play a role, the role is just hidden.
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$\begingroup$ Formula states that ω depends on k and m. So, without changing them ω remains constant? Or it depends on some other factors as well? $\endgroup$ Commented Oct 12, 2017 at 19:52
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$\begingroup$ The frequency of the first SHM (mass on a spring) depends only on k and m, so yes, if you want an oscillator with a different $\omega$, you must change one of those two. $\endgroup$– PaulWCommented Oct 12, 2017 at 20:14
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$\begingroup$ So we can say that energy does not depend on ω, rather m and k. $\endgroup$ Commented Oct 12, 2017 at 20:18
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$\begingroup$ I would say energy does depend on frequency. Because if you have two different SHOs with different frequencies but the same mass and amplitude, the energies will be different. They also must have different spring constants, but that doesn't mean energy has no relation to frequency. For there to be no relation, I would say a graph of energy vs. frequency would have to be a flat line (zero slope) and this is not the case. $\endgroup$– PaulWCommented Oct 12, 2017 at 20:27
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$\begingroup$ It's dangerous to say that energy depends on frequency. I don't like it one bit. You are not free to change the frequency. If you are given a mass and a spring, you have the attributes of those objects. The spring constant and mass characterize the system. The frequency is then fixed. The frequency follows from $k$ and $m$. $\endgroup$– garypCommented Feb 25, 2018 at 19:26
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The other way of looking at this is considering the work done by spring and the energy as the area under the curve of force multiplied by max displacement of mass.
Obviously this is a triangle witch passes the origin with a slope of K. Hence the total area of the triangle of work is $1/2 K.X^2 $ Because at the beginning when displacement is small work done by the system is small and it grows linearly as displacement grows until spring is loaded to K.X.
As has been mentioned in other answers $\omega$ is related to K and M but it will obligingly fall into place respecting our calculation!
