I have this question with the answer listed as $2.0\,\mathrm{m/s}$.

"A $1.25\,\mathrm{kg}$ mass on a spring with a constant of $12.0\,\mathrm{N/m}$ is oscillating back and forth. Its maximum speed is $2.5\,\mathrm{m/s}$. What speed will it have when it is $0.48\,\mathrm{m}$ from equilibrium?"

But I'm struggling to understand that.

What I did was $mv^2 = kx^2$, which eventually translates down to $0.48\,\sqrt{12/1.25} \simeq 1.487\,\mathrm{m/s}$.

Is the answer just a simple mistake, or am I miscalculating this?


closed as off-topic by John Rennie, Thomas Fritsch, David Z Jun 25 at 7:19

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  • $\begingroup$ Why do you think the kinetic energy would be equal to the elastic potential energy? Rather, they will add to a constant value. You could use $\Delta E_k=-\Delta E_e$ $\endgroup$ – M. Enns Jun 25 at 1:37
  • $\begingroup$ I added MathJax, please use it in the future for typesetting math. See guide and links therein. I've also added the tag "Homework and exercises." Please use it in the future for similar questions. Also check out the meta and the guide. $\endgroup$ – MannyC Jun 25 at 4:54
  • $\begingroup$ Hi Jason and welcome to the Physics SE! Please note that we don't answer homework or worked example type questions. Please see this page in the site help for more on what topics you can ask about here. $\endgroup$ – John Rennie Jun 25 at 5:03
  • $\begingroup$ The idea to use the energy conservation is right, you just need to be careful about which $x$ and $v$ you use (by which I mean at which time). You need however to use additional relations between the dynamics of the position and the speed to find the correct answer. If it reassures you, I find 2 $m/s$. $\endgroup$ – David Jun 25 at 9:30

Your formula mv^2 = kx^2 is baseless. You should write the energy conservation law correctly.


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