Question

One end of a spring of negligible un-stretched length and spring constant k is fixed at the origin $(0,0)$. A point particle of mass m carrying a positive charge q is attached at its other end. The entire system is kept on a smooth horizontal surface. When a point dipole p​ pointing towards the charge q is fixed at the origin, the spring gets stretched to a length l and attains a new equilibrium position (see figure below). If the point mass is now displaced slightly by $\Delta l < < l$ from its equilibrium position and released, it is found to oscillate at frequency $\dfrac{1}{\delta }\sqrt {\dfrac{k}{m}} $ The value of $\delta $ is_______

Answer 1

Hint: In order to solve this question, we will find the net potential energy of the spring and dipole system and then we will find the equilibrium point of the system and then using general frequency formula of a variable system we will find the value of $\delta $.

Formula used:
If U is the potential energy of the system varying with distance x with mass m then frequency of the system is given by $f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{(\dfrac{{{d^2}U}}{{d{x^2}}})}}{m}} $
Potential energy due to dipole P and charge q is given by $P.E = \dfrac{{p \times q}}{{4\pi {\varepsilon _0}{x^2}}}$

Complete answer:
According to the question, let the displacement by the spring is x when dipole came in to act, then potential energy stored in the spring will be $\dfrac{1}{2}k{x^2}$ and potential energy due to dipole and charge is $P.E = \dfrac{{p \times q}}{{4\pi {\varepsilon _0}{x^2}}}$ so, net potential energy of the system will be
$U = \dfrac{1}{2}k{x^2} + \dfrac{{p \times q}}{{4\pi {\varepsilon _0}{x^2}}}$
Now, To find equilibrium point we have
$\dfrac{{dU}}{{dx}} = 0$
$ \Rightarrow kx - \dfrac{{2pq}}{{4\pi {\varepsilon _0}{x^3}}} = 0$
now, the equilibrium position of the system is at $x = l$ so on putting the value, we get
$
  kl - \dfrac{{2pq}}{{4\pi {\varepsilon _0}{l^3}}} = 0 \\
  kl = \dfrac{{2pq}}{{4\pi {\varepsilon _0}{l^3}}} \to (i) \\
 $
Now,
$\dfrac{{{d^2}U}}{{d{x^2}}} = k + \dfrac{{6pq}}{{4\pi {\varepsilon _0}{l^4}}}$
put the value of $kl = \dfrac{{2pq}}{{4\pi {\varepsilon _0}{l^3}}}$ from equation (i) we get,
$
  \dfrac{{{d^2}U}}{{d{x^2}}} = k + 3k \\
  \dfrac{{{d^2}U}}{{d{x^2}}} = 4k \\
 $
Now, let f be the frequency of the system then its frequency is given by
$f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{(\dfrac{{{d^2}U}}{{d{x^2}}})}}{m}} $ on putting the values we get
$
  f = \dfrac{1}{{2\pi }}\sqrt {\dfrac{{4k}}{m}} \\
  f = \dfrac{1}{\pi }\sqrt {\dfrac{{2k}}{m}} \\
 $
on comparing this value with $\dfrac{1}{\delta }\sqrt {\dfrac{k}{m}} $ we get,
$
  \delta = \pi \\
  \delta = 3.14 \\
 $
Hence, the value of $\delta = 3.14$

Note: Generally, a stretched spring has potential energy stored in itself and when we release it, the potential energy will be converted into its kinetic energy. Therefore the work done of that spring is proportional to the square of the displacement.