Testing paint skills( physics question)

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

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

Image text Image text

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

Note by Jatin Yadav
5 years, 10 months ago

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12 votes

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Hi Jatin!

Here's my solution:

At equilibrium,

mg=λ22πϵ0ln(h+lh)\displaystyle mg=\frac{\lambda^2}{2\pi \epsilon_0}\ln\left(\frac{h+l}{h}\right)

Lets displace the red wire vertically upwards by yy.

The net force acting on the red wire is:

Fnet=λ22πϵ0ln(1+lh+y)mg\displaystyle F_{net}=\frac{\lambda^2}{2\pi\epsilon_0}\ln\left(1+\frac{l}{h+y}\right)-mg

Using the approximation,

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

Fnet=λ2ly2πh(l+h)ϵ0y¨=λ2l2mπh(l+h)ϵ0y\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! :)

Pranav Arora - 5 years, 10 months ago

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It was correct!, but you missed mm in the denominator, i do these approximations as:

d(ln(1+lh))=11+lh×lh2dhd(\ln (1 +\frac{l}{h})) = \frac{1}{1 + \frac{l}{h}} \times \frac{- l}{h^2} dh , where dh=ydh = y, dd representing small change.

jatin yadav - 5 years, 10 months ago

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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! :)

Pranav Arora - 5 years, 10 months ago

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@Pranav Arora It was something where we had to apply gauss law in mirror problem, but since, i was a beginner that time in making problems i later noted that it had some problem, So, i had to delete it to avoid confusion, but still i can at least e-mail it at your id, which you might tell.

jatin yadav - 5 years, 10 months ago

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@Jatin Yadav Ah, I don't like to post my address in public but I remember you posted your e-mail id in one of the discussions. I remember your e-mail address so I will send you a mail. :)

Pranav Arora - 5 years, 10 months ago

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You should also have shown how force of interaction is λ22πϵ0ln(h+lh)\frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg), I do it here,

Consider a small element of length dxdx having charge dq=λdxd q = \lambda dx at a distance xx from blue wire, we know that

Ex=λ2πϵ0xE_{x} = \frac{\lambda}{2 \pi \epsilon_{0} x}

F=hh+lExλdx=hh+lλ2πϵ0xλdxF = \displaystyle \int_{h}^{h+l} E_{x} \lambda dx = \int_{h}^{h+l} \frac{\lambda}{2 \pi \epsilon_{0} x} \lambda dx

= λ22πϵ0ln(h+lh) \frac{\lambda^2}{2 \pi \epsilon_{0}} \ln\bigg(\frac{h + l}{h}\bigg)

jatin yadav - 5 years, 10 months ago

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How did you approximate it ??

Priyanka Banerjee - 5 years, 10 months ago

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lambda/root(8pie^2epsilon zero*(h+l))...I think this is the answer..if solution is needed I will provide one later..a lil' bit busy!

Susovan Maity - 5 years, 10 months ago

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No, this is not the answer.

jatin yadav - 5 years, 10 months ago

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

Susovan Maity - 5 years, 10 months ago

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Challenge

Solve this within a minute:

We remove the blue wire and and take the red wire to a place where, electric field varies as E=E0ex2E =E_{0} e^{-x^2} It stays in equilibrium at some height HH say,and i repeat the same experiment, Task remains same , find the frequency of oscillations.

jatin yadav - 5 years, 10 months ago

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S I unit of force

Vinay Chandra - 5 years, 5 months ago

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