- #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.

- Thread starter oddjobmj
- Start date

- #1

- 306

- 0

Please delete this post.

Last edited:

- #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!

Share:

- Replies
- 5

- Views
- 5K