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Dunkodx

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Thread moved from the technical forums, so no Homework Template is shown

**Summary::**Doubt in a spring exercise

Text of the exercise "a mass of ##m = 0.4 \ \text{kg} ## is attached to a spring and it oscillates horizontally with period ##T = 1.57 \text{s}##; the amplitude of the oscillation is ##d = 0.4 \text{m}##. Determine the spring constant, the total energy of the system and the position where the kinetic energy is equal to the potential energy."

I have found the spring constant with the relation ##T = \frac{2\pi}{\omega}## and I've used the conservation of energy to say that ##E=\frac{1}{2} k \left(\frac{d}{2}\right)^2=\frac{1}{2}mv^2=\frac{1}{2} m\omega^2 \left(\frac{d}{2}\right)^2##; I have a doubt for the last request, that is the position where it is ##E_{\text{k}}=U##; my reasoning is that in a generic position it is, again for the conservation of energy, that ##\frac{1}{2}kx^2+\frac{1}{2}mv^2=\frac{1}{2}k \left(\frac{d}{2}\right)^2##, but since we want the position where ##E_{\text{k}}=U## and it is ##E=E_{\text{k}}+U## substituting ##E_{\text{k}}=U## leads to ##E_m=2U \implies E_{\text{k}}+U= 2\cdot \frac{1}{2} k \left(\frac{d}{2}\right)^2\implies \frac{1}{2}kx^2+\frac{1}{2}mv^2=k \left(\frac{d}{2}\right)^2##. Again, since ##\frac{1}{2}kx^2=\frac{1}{2}mv^2## because I am interested when kinetical and potential energy are the same, I get ##kx^2=k\left(\frac{d}{2}\right)^2 \implies x=\frac{d}{2}##.

However the solution says that ##x=\frac{d}{2\sqrt{2}}##, where is my mistake? Thanks.