# Find the speed of the particle when it has covered a displacement of magnitude $L$ [closed]

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.

• 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. Commented Oct 8, 2022 at 18:45

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.
• I got $x=\frac{3\lambda}{4}$ but even after taking this limit my answer is not coming right...pls help Commented Oct 9, 2022 at 4:59