-1
$\begingroup$

This is my first question on this site.

A particle of mass $m$ is constrained to move along $x-\textrm{axis}$. The only force acting on the particle is given by $$F=F_0\cos\left(2\pi\frac{x}{\lambda}\right)$$ where $x$ is the position of the particle on the $x-\textrm{axis}$. It is at rest on one of the equilibrium position but the equilibrium is unstable. Due to slight disturbance it moves, write the speed of the particle when it has covered a displacement of magnitude $L$.

My work:

I discovered the acceleration as a function of $x$ by dividing the force by $m$. Then I used the fact that $$a=v\frac{\textrm{d}v}{\textrm{d}x}$$ and integrated with the lower limit as $0$ and the upper limit as $L$. After all the simplifications my answer came $$\sqrt{\frac{F\lambda}{m\pi}\sin\left(\frac{2\pi L}{\lambda}\right)}$$ but the correct answer given is $$\sqrt{\frac{F\lambda}{m\pi}\left(1-\cos\left(\frac{2\pi L}{\lambda}\right)\right)}$$ Where am I wrong$?$ Do I have to take the limits as $\frac{3\pi}{2}$ and $\frac{3\pi}{2}+L?$

Any help is greatly appreciated.

I think that those limits are right as the particle is in unstable equilibrium and at $\frac{3\pi}{2}$ by the graph of cosine function, the particle is at unstable equilibrium.

$\endgroup$
1
  • 1
    $\begingroup$ Check my work questions are off-topic here. To make your question on-topic, you need to edit it to ask about the concepts underlying your calculations, and state why you think those equations & limits are appropriate. $\endgroup$
    – PM 2Ring
    Commented Oct 8, 2022 at 18:45

1 Answer 1

1
$\begingroup$

You have done everything right, your mistake is that: while at $\frac{3\pi}{2}$, the particle is indeed at unstable equilibrium, it is not the value of $x$, but the argument of cosine that is $\frac{3\pi}{2}$. Therefore, $$\frac{2\pi x}{\lambda}=2n\pi +\frac{3\pi}{2} $$where $n\in \mathbb{Z}$

This gives the set of all points of unstable equilibrium. From there you can carry out the integration from $x$ equal to any of the above points, to "$L$ plus that $x$", to give you the correct speed.

$\endgroup$
2
  • $\begingroup$ I got $x=\frac{3\lambda}{4}$ but even after taking this limit my answer is not coming right...pls help $\endgroup$
    – Vanessa
    Commented Oct 9, 2022 at 4:59
  • $\begingroup$ It came thanks very much $\endgroup$
    – Vanessa
    Commented Oct 9, 2022 at 6:02

Not the answer you're looking for? Browse other questions tagged or ask your own question.