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.
[Physics] Does total energy in SHM depend on frequency
frequencyharmonic-oscillator
Related Solutions
Because the frequency of a sound wave is defined as "the number of waves per second."
If you had a sound source emitting, say, 200 waves per second, and your ear (inside a different medium) received only 150 waves per second, the remaining waves 50 waves per second would have to pile up somewhere — presumably, at the interface between the two media.
After, say, a minute of playing the sound, there would already be 60 × 50 = 3,000 delayed waves piled up at the interface, waiting for their turn to enter the new medium. If you stopped the sound at that point, it would still take 20 more seconds for all those piled-up waves to get into the new medium, at 150 waves per second. Thus, your ear, inside the different medium, would continue to hear the sound for 20 more seconds after it had already stopped.
We don't observe sound piling up at the boundaries of different media like that. (It would be kind of convenient if it did, since we could use such an effect for easy sound recording, without having to bother with microphones and record discs / digital storage. But alas, it just doesn't happen.) Thus, it appears that, in the real world, the frequency of sound doesn't change between media.
Besides, imagine that you switched the media around: now the sound source would be emitting 150 waves per second, inside the "low-frequency" medium, and your ear would receive 200 waves per second inside the "high-frequency" medium. Where would the extra 50 waves per second come from? The future? Or would they just magically appear from nowhere?
All that said, there are physical processes that can change the frequency of sound, or at least introduce some new frequencies. For example, there are materials that can interact with a sound wave and change its shape, distorting it so that an originally pure single-frequency sound wave acquires overtones at higher frequencies.
These are not, however, the same kinds of continuous shifts as you'd observe with wavelength, when moving from one medium to another with a different speed of sound. Rather, the overtones introduced this way are generally multiples (or simple fractions) of the original frequency: you can easily obtain overtones at two or three or four times the original frequency, but not at, say, 1.018 times the original frequency. This is because they're not really changing the rate at which the waves cycle, but rather the shape of each individual wave (which can be viewed as converting some of each original wave into new waves with two/three/etc. times the original frequency).
Let's start by plotting the damped harmonic oscillation equation:
$$x = Ae^{-\gamma t} \cos (\omega t)$$
In the above example $\gamma = 0.8$ and $\omega = 2.5\times\pi$. It should now be more clear that when $t = 0$, $x=A$ i.e. the initial displacement of the oscillator from it's equilibrium position. Here $A = 1~\mathrm{m}$. Let's assume down is the positive $x$-axis for simplicity.
Now the way one would conventionally go about finding the damping coefficient is by finding $\gamma$ form the envelope function $Ae^{-\gamma t}$ (plotted in green). You would ideally use a data logger of some fashion to sample the position of the oscillator as a function of time (hopefully with enough resolution) and then you find $x$ at the exact same point in the cycle (constant phase) for each oscillation. It is usually the easiest to locate the peaks (plotted in red) where $$\cos(\omega t) = 1 \quad \therefore \quad \omega t = 2n\pi\; (n=1,2,3,\ldots) $$
You can either fit an exponential curve to these data points or alternatively fit a linear curve to the log-log of the data.
Now in your case, it seems you only have 2 data points. The position at $t =0$, and the time $t_{1/2}$ when $x= A/2$. It's not ideal, but you have everything you need to calculate the damping coefficient now.
Hope this helps :)
Best Answer
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.