Finding Spring Constant & Energy w/ Doubt in Exercise

[Dunkodx]

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.

 

[hutchphd]

What is $E_m$ in your equation soup above?

 

[Dunkodx]

@hutchphd: It was a typo, that $E_m$ is the same as all the other $E$ in the post. Sorry for the confusion.

 

[hutchphd]

O.K. Why do you say $E=\frac1 2 k (\frac d 2 )^2$

 

[Dunkodx]

Because there are only conservative forces, so when the mass reaches the maximum amplitude (that is, when it is at position $\frac{d}{2}$) the kinetical energy is $0$ because the mass has no velocity and there is only potential energy $\frac{1}{2} k \left(\frac{d}{2}\right)^2$; that means that $E=\frac{1}{2}k \left(\frac{d}{2}\right)^2$.

 

[hutchphd]

The amplitude is the size of the excursion from equilibrium in my vernacular. Check this.

 

[nasu]

The question explicitly states that the amplitude is d. And there is no "maximum amplitude". Amplitude is the maximum displacement from equilibrium. Unless otherwise stated (like in peak-to-peak amplitude).

 

[Steve4Physics]

Because there are only conservative forces, so when the mass reaches the maximum amplitude (that is, when it is at position $\frac{d}{2}$) the kinetical energy is $0$ because the mass has no velocity and there is only potential energy $\frac{1}{2} k \left(\frac{d}{2}\right)^2$; that means that $E=\frac{1}{2}k \left(\frac{d}{2}\right)^2$.

If the equilibrium position is x=0, then displacement (x) is in the range $-d≤x≤d$.
Extremum-to-extremum distance is 2d (twice the amplitude).
That means maximum potential energy is $\frac 1 2 kd^2$, not $\frac 1 2 k(\frac d 2)^2$.