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Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Consider the following macroscopic oscillator: two masses of equal mass $m$, are attached to a spring and oscillate with amplitude $A$, at a frequency $v$.

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. 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).

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. 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). Not the answer you're looking for?Browse other questions tagged homework-and-exercises spring classical-mechanics or ask your own question.

Why does Uncle Bob suggest that coding standards shouldn't be written down if you can avoid it? Returns & Refund Guarantee details Buyers can receive a refund and keep the item(s) once the item(s) are not as described or possess any quality issues by negotiating directly with the seller. Note: Due to possible delay of exchange rate update, price in various currencies is for reference only. 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 $t$: $V = \frac{1}{2}k{l^2}\left( {{x_{1,}}{x_2}} \right) = \frac{1}{2}k\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right)$. 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. At any other time, the potential energy is less than this and the kinetic energy is greater so that the sum remains constant. Leather cheap branded watches are luxurious and look expensive, plastic buy cheap watches online are cute and can be in multiple colors, while metal buy cheap watches are tough and durable for years.

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 $E$. As for the kinetic energy of the two masses, we have $T = \frac{1}{2}m\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right)$. The maximum will occur every time $\cos^2 \left( {\omega t + \phi } \right) = 1$, wich will happen twice in every period of the oscillation. 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).

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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. 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).

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. 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). Not the answer you're looking for?Browse other questions tagged homework-and-exercises spring classical-mechanics or ask your own question.

Why does Uncle Bob suggest that coding standards shouldn't be written down if you can avoid it? Returns & Refund Guarantee details Buyers can receive a refund and keep the item(s) once the item(s) are not as described or possess any quality issues by negotiating directly with the seller. Note: Due to possible delay of exchange rate update, price in various currencies is for reference only. 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 $t$: $V = \frac{1}{2}k{l^2}\left( {{x_{1,}}{x_2}} \right) = \frac{1}{2}k\left( {{x_1}^2 + {x_2}^2 + 2{x_1}{x_2}} \right)$. 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. At any other time, the potential energy is less than this and the kinetic energy is greater so that the sum remains constant. Leather cheap branded watches are luxurious and look expensive, plastic buy cheap watches online are cute and can be in multiple colors, while metal buy cheap watches are tough and durable for years.

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 $E$. As for the kinetic energy of the two masses, we have $T = \frac{1}{2}m\left( {{{\dot x}_1}^2 + {{\dot x}_2}^2} \right)$. The maximum will occur every time $\cos^2 \left( {\omega t + \phi } \right) = 1$, wich will happen twice in every period of the oscillation. 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).

The cheap wrist watches in our shop are precise and delicate, you can never be disappointed with them, and they can totally become your proud on the wrist!

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