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The amplitude of the resulting oscillation is $\frac{xqQ}{8\pi ^2 \epsilon _0 a^3 k}$ .Find the value of 'x'?

2 votes

In the given figure ABC is a non-conducting semi-circular wire of
radius 'a' carrying a total charge Q uniformly distributed on it and a
point charge q is at its centre. Ends of the wire are attached to two
separate springs each having spring constant k as shown in the
figure. In the given position system is in equilibrium. Now the point
charge is suddenly removed. The amplitude of the resulting oscillation
is $\frac{xqQ}{8\pi ^2 \epsilon _0 a^3 k}$ . Ignoring any force other than spring force and
electrostatic force find the value of 'x'?

My try :

asked Aug 24, 2018 in Physics Problems by koolman (4,236 points)

1 Answer

2 votes
Best answer

Your calculation of $T$ is correct - and possibly simpler than integration, which is the usual method for finding the electrostatic repulsion force $F$. Perhaps your difficulty is that the result which you get for the amplitude does not match the answer which you have been given for $x$?

The wire is initially in equilibrium between two forces : the repulsive force $F$ between the two charges $+Q$ and $+q$ which pushes the wire to the right, and the 2 tensions $T$ in each spring which pull the wire to the left : $F=2T$. The initial extension of the springs is the amplitude of oscillations : $T=kA$.

You are not asked to find the period of oscillations, so it is not necessary to write the equation of motion and deduce the angular frequency. To find the period we would need to know the mass of the wire, which is not stated. The amplitude will be the same regardless of the mass of the wire.

You have found $T$. Finding the amplitude $A$ is then straightforward. If the result $x=2a$ is not what you expected, there may be an error in the question or in the given answer.

$qQ/4\pi \epsilon_0 a^2$ has the unit of force ($N$), and $k$ has unit of force per length ($N/m$). So in order for $xqQ/8\pi^2\epsilon_0 a^3 k$ to have the unit of length (amplitude) then $x$ must also have the unit of length.

Note : Do not confuse $x$ with the extension of the springs. In this question $x$ is an unknown factor in an equation. The initial extension of the springs is the amplitude $A$.

answered Aug 24, 2018 by sammy gerbil (27,556 points)
selected Mar 12 by koolman
Sorry I know this is not my question but I saw the method koolman applied and I do not understand it. First of all why $Q = \pi a$ and not $Q = \sigma A$? Where $\sigma$ is the surface charge density. Also I have been trying to obtain in a geometrically way $2Tsin(\frac{d\theta}{2})$ and got $T = a sin(\frac{d\theta}{2})$ instead (which has to be wrong but geometrically  speaking I got it).
Koolman does not say $Q=\pi a$.  He says that the charge on a small segment of the wire which subtends angle $d\theta$ radians is a fraction $\frac{d\theta}{\pi}$ of the total charge $Q$ on the wire, because the total angle subtended by the wire is $\pi$. There is no surface density because the wire is a line with only 1 dimension, so it has no surface area.

Koolman's calculation says that the tension $T$ in the wire is constant. On a small element of wire at B which subtends angle $d\theta$ the 2 forces $T$ which pull on it are not aligned. Each is inclined at angle $\frac12 d\theta$ to the vertical. The vertical components $T\cos(\frac12 d\theta)$ cancel out but the horizontal components $T\sin(\frac12 d\theta)$ add. So there is a resultant force of $2T\sin(\frac12d\theta)$ acting towards $q$. This opposes the electrostatic repulsion from $q$ which acts on the charge on this small segment of wire.
Thanks I understood it :) By the way I also got $x=2a$
Although I got my answer but still I was thinking that we can find the equation of the SHM
The equation of motion is easy. This is just a mass on a spring. The equation is $y=A\cos\omega t$ where amplitude $A$ is what you found and $\omega^2=2k/m$ with $m$ being the mass of the wire, which is not given.