Hi , i made this question yesterday, and it follows :

Consider an infinitely large wire (blue) having charge per unit length \(+ \lambda\), and a finite length wire (red) of length \(l\) and mass \(m\) having the same charge per unit length. Its lower end rests at a height of \(h\) units from the blue wire,as shown in the figure.

Your task is simple. Find the frequency of oscillation of red wire on being disturbed by a small distance vertically

No vote yet

12 votes

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestHi Jatin!

Here's my solution:

At equilibrium,

\(\displaystyle mg=\frac{\lambda^2}{2\pi \epsilon_0}\ln\left(\frac{h+l}{h}\right)\)

Lets displace the red wire vertically upwards by \(y\).

The net force acting on the red wire is:

\(\displaystyle F_{net}=\frac{\lambda^2}{2\pi\epsilon_0}\ln\left(1+\frac{l}{h+y}\right)-mg\)

Using the approximation,

\(\displaystyle \ln\left(1+\frac{l}{h+y}\right)\approx \ln\left(\frac{h+l}{h}\right)-\frac{ly}{h(l+h)} \)

\(\displaystyle F_{net}=-\frac{\lambda^2ly}{2\pi h(l+h)\epsilon_0} \Rightarrow \ddot{y}=-\frac{\lambda^2l}{2m\pi h(l+h)\epsilon_0}y\)

The above equation is for SHM and the frequency can be easily deduced.

Is this correct?

I have a small question, how do you take such approximations? I had to use Wolfram Alpha, can you provide some help on this? Many thanks! :)

Log in to reply

It was correct!, but you missed \(m\) in the denominator, i do these approximations as:

\(d(\ln (1 +\frac{l}{h})) = \frac{1}{1 + \frac{l}{h}} \times \frac{- l}{h^2} dh \), where \(dh = y\), \(d\) representing small change.

Log in to reply

Thanks Jatin! Yes, I missed the m, sorry about that, I will edit that. :)

I was wondering if you don't mind, could you please repost the same physics problem about force on mirror you posted before, I am very interested to know about its solution. Thanks! :)

Log in to reply

Log in to reply

Log in to reply

You should also have shown how force of interaction is \(\frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg)\), I do it here,

Consider a small element of length \(dx\) having charge \(d q = \lambda dx\) at a distance \(x\) from blue wire, we know that

\(E_{x} = \frac{\lambda}{2 \pi \epsilon_{0} x}\)

\(F = \displaystyle \int_{h}^{h+l} E_{x} \lambda dx = \int_{h}^{h+l} \frac{\lambda}{2 \pi \epsilon_{0} x} \lambda dx\)

= \( \frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg)\)

Log in to reply

How did you approximate it ??

Log in to reply

lambda/root(8

pie^2epsilon zero*(h+l))...I think this is the answer..if solution is needed I will provide one later..a lil' bit busy!Log in to reply

No, this is not the answer.

Log in to reply

may have done something wrong during the approximation..sorry!

Log in to reply

ChallengeSolve this within a minute:

We remove the blue wire and and take the red wire to a place where, electric field varies as \(E =E_{0} e^{-x^2}\) It stays in equilibrium at some height \(H\) say,and i repeat the same experiment, Task remains same , find the frequency of oscillations.

Log in to reply

S I unit of force

Log in to reply