- #1

- 306

- 0

Please delete this post.

Last edited:

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

In summary, you are using the wrong equation for the force, and the correct equation is du/dx= -f(x). You also need to use Newton's second law, not potential energy. Finally, you can find x as a function of t using a sinusoidal oscillation with period 2 pi sqrt(frac m(2c))

- #1

- 306

- 0

Please delete this post.

Last edited:

Physics news on Phys.org

- #2

- 28

- 3

du/dx=f(x)..(this is where you are making mistake).The correct equation is du/dx= -f(x)

so f(x)= -2cx(This force is opposite to the direction of velocity of particle.i.e.the force is in negative x-direction..

Now do the process again and you will easily reach the answer..

- #3

- 6,054

- 391

You already have conservation of energy: ## \frac {mv^2} {2} + cx^2 = \frac {mv_0^2} {2} ##, thus ## v = \sqrt { v_0^2 - 2 \frac c m x^2} ##, or ## \frac {dx} {dt} = \sqrt { v_0^2 - 2 \frac c m x^2} ##.

Can you solve this?

If you insist on using Newton's second law, then I should say that this step of yours: ## \int 2cxdt = \int m dv \Rightarrow \frac {2cxt} m = v + \text{constant} ## is already wrong, and not really because of the missing minus sign, but because ##x## is a function of time, no a constant, so its integral is not ##xt##.

- #4

- 6,054

- 391

\int_0^{x_m} \frac {dx} {\sqrt{v_0^2 - 2 \frac c m x^2} } = \int dt

\\

\Rightarrow

\int_0^{x_m} \frac {dx} {\sqrt{1 - (ax)^2} } = v_0 \int dt

$$ where $$

a^2 = \frac {2c} {mv_0^2}

$$ Then $$

\frac a a \int_0^{x_m} \frac {dx} {\sqrt{1 - (ax)^2} } = \frac 1 a \int_0^{ax_m} \frac {d(ax)} {\sqrt{1 - (ax)^2} } = \frac 1 a (\arcsin ax_m - \arcsin 0) = \frac 1 a \arcsin ax_m = v_0 t

$$ Now ## x_m = \sqrt { \frac {mv_0^2} {2c} } = a^{-1} ##, so ##\frac 1 a \frac {\pi} 2 = v_0 t ## giving $$ t = \sqrt {\frac m {2c} } \frac {\pi} 2 $$ Also of note is that you can find ##x## as a function of ##t##: $$ x = a^{-1} \sin av_0t $$ which is a sinusoidal oscillation with period ## \frac {2 \pi} {av_0} = 2 \pi \sqrt { \frac {m} {2c}} ##. Clearly the time it takes from max velocity to max displacement is one quarter of the period, which is consistent with the previous result.

If you have studied linear differential equations, you could also observe that Newton's second law gives $$ \frac {d^2 x} {dt^2} + \frac {2 c} {m} x = 0 $$ which is a second-degree linear diff. eq., whose general solution is $$ x = A \sin (\sqrt{\frac {2c} m}t + \alpha) $$ which is again periodical with the same period, of course.

- #5

- 306

- 0

Thanks again!

The maximum x value for a particle can be found by taking the derivative of the potential function and setting it equal to zero. Solving for x will give the position of the particle at which the potential is at its maximum.

At the maximum x value, the particle will come to a stop and then begin to move in the opposite direction due to the force from the potential function.

The shape of the potential function can greatly affect the maximum x value. A steep potential function will result in a smaller maximum x value, while a flatter potential function can lead to a larger maximum x value.

No, the maximum x value is the point at which the particle's motion will reverse. It cannot travel beyond this point without a change in the potential function or an external force acting on the particle.

Yes, there are other factors that can affect the maximum x value, such as the initial velocity of the particle and any external forces acting on it. These factors can change the overall motion of the particle and therefore impact the maximum x value.

Share:

- Replies
- 13

- Views
- 526

- Replies
- 4

- Views
- 910

- Replies
- 25

- Views
- 845

- Replies
- 14

- Views
- 843

- Replies
- 10

- Views
- 782

- Replies
- 6

- Views
- 564

- Replies
- 6

- Views
- 348

- Replies
- 4

- Views
- 375

- Replies
- 4

- Views
- 776

- Replies
- 26

- Views
- 1K