# Equal mass on either end of a spring

I'm taking physics this term and this is the first time taking physics in 3 years. Here is my homework question (of course those variables all have values but we'll use variables for now):

Consider the following macroscopic oscillator: two masses of equal mass $m$, are attached to a spring and oscillate with amplitude $A$, at a frequency $v$. Use classical physics to calculate the energy of the oscillator.

This is how I did it. Can someone tell me if my solution is correct?

Since the question didn't say where the masses are and under what environment, I assumed they are on either end of the spring and the environment is frictionless. Let $k$ be the spring constant. For each of the blocks, we have the differential equation $m\frac{d^2x}{dt^2}=-k(2x)$. This has the solution $x(t)=A\cos(\sqrt{\frac{2k}{m}}t)$. Therefore, the frequency is $\frac{\sqrt{\frac{2k}{m}}}{2\pi}$. Equating this to $v$, we can find the value of $k$. Hence, the energy of the oscillator is $\frac{1}{2}k(2A)^2$.

Edit: My main concern is my computation of $k$. Because I tried to look for a similar question and I found http://www.uccs.edu/~rtirado/Ch14%20ISM.pdf problem 99. Their computation does not involve $2\pi$ anywhere. Can someone possibly explain that to me?

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Do you know answer? A new answer has come. – ABC May 26 '13 at 2:44
Please use the homework tag for homework questions. – Ben Crowell May 26 '13 at 3:08

Your answer $E=2kA^2$ is correct for the total energy of the system (kinetic energy of the two masses and elastic energy of the oscillator). In this problem, the sum of such energies is constant according to the law of conservation of energy, so that $E=T+V$. This means that the energy of the oscillator will increase and decrease at times so that when added to the kinetic energy their sum remains constant and equal to $E$.

My approach to this problem would be to start by writing the potential energy of the oscillator as a function of the coordinates of the 1-dimensional problem. Let $l$ be the elongation of the oscillator, $l(x_1,x_2)=x_2-x_1$, and $x_i$ the positions of the masses at each end of the oscillator at any time $t$: $V = \frac{1}{2}k{l^2}\left( {{x_{1,}}{x_2}} \right) = \frac{1}{2}k\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right)$.

Here $x_i(t)=A_i\cos(\omega_it+\phi_i)$ and $\omega = \sqrt{\frac{k}{m}}$. In this problem one can figure out that both masses oscillate at the same frequency and with the same amplitude if given the same initial conditions. As for the kinetic energy of the two masses, we have $T = \frac{1}{2}m\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right)$.

As we know, $E = T + V = {\rm{const.}}$ and It's trivial that the potential energy of the oscillator will "oscillate" between 0 (maximum kinetic energy) and some value $V_\rm{max}$ (zero kinetic energy). The maximum potential energy will be equal to the total energy of the system. The maximum will occur every time $\cos^2 \left( {\omega t + \phi } \right) = 1$, wich will happen twice in every period of the oscillation.

Therefore $E=V_{\rm max}=\frac{1}{2}k\left( {{A^2} + {A^2} + 2{A^2}} \right) = 2k{A^2}$. At any other time, the potential energy is less than this and the kinetic energy is greater so that the sum remains constant. If you made a plot of both energies it'd look something like this:

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Did you use gnuplot to plot the graph? – sayantankhan May 26 '13 at 13:27
And what is the value of $k$? Also, $l^2(x_1,x_2)=x_1^2+x_2^2-2x_1x_2$ Also, how did you deduce what $x_i(t)$ is? – user44322 May 26 '13 at 17:12
$\omega = 2\pi \nu = \sqrt {\frac{k}{m}} \Rightarrow k = 4{\pi ^2}{\nu ^2}m$. $l(x_1,x_2)=x_2-x_1$ which is the distance between the two masses in the $x$-dimension. To deduce $x_i(t)$ you have to solve the dynamic problem, that is, the differential equation of motion that results from the Euler-Lagrange equations (or Newton's second law in a less detailed course). – Mike D. May 26 '13 at 18:36
The graph was plotted with gnuplot. – Mike D. May 26 '13 at 18:38

I think as nothing explanatory is given: $A$ is amplitude of 1 mass or both have seperate amplitudes A/2 which cause net extention as $A$.

We should do this by equivalent mass. Equivalent/Reduced mass of system is $$(m*m)/(m+m)=m/2$$

So, $$w=\sqrt{K/(m/2)}$$ we'll get frequency and $K$

So, energy is $$E=kA^2/2$$

Maybe it's wrong....

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+1 for new user! – ABC May 26 '13 at 2:52
Why is it over $4$? – user44322 May 26 '13 at 3:48
@user44322 Sorry it's 2. – xyzabc May 26 '13 at 7:32
@userØØ7 thank u – xyzabc May 26 '13 at 7:32