Question

In the figure shown A and B are two charged particles having charges $q$ and $-q$ respectively are placed on a nonconducting fixed horizontal smooth plane. B is fixed and A is attached to a non conducting massless spring of spring constant $K$. The other end of the spring is fixed. Mass of A is $m$, A and B are in equilibrium, when the distance between them is $r$. Choose the correct options

• A. time period of small oscillation of block $A$ about is mean position $=2\pi \sqrt{\frac{m}{K-\frac{q^2}{2\pi {ϵ}_0{r}^3}}}$
• B. time period of small oscillation of block A about is means position $=2\pi \sqrt{\frac{m}{K+\frac{q^2}{2\pi {ϵ}_0{r}^3}}}$
• C. to perform SHM, $K$ must be greater than
  $\frac{q^2}{2\pi {ϵ}_0{r}^3}$
• D. Will perform SHM, for any value of $K$

Answer 1

Answer: (A), (C)

to perform SHM, $K$ must be greater than

$\frac{q^2}{2\pi {ϵ}_0{r}^3}$
In equilibrium if elongation in spring is $x_0$
$F_s=F_e$
$\Rightarrow K{x}_0=\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}\dots \left(1\right)$

At any instant ,if A is displaced by $x$ towards right, FBD can be considered as
Here
$F_e=\frac{1}{4\pi {ϵ}_0}$ $\frac{q^2}{(r-x{)}^2}$
Hence
$a=\frac{\left[K\right(x+x_0)-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{(r-x{)}^2}]}{m}$
using binomial theorem, as $x<<r$ and equation 1
we get
$a=\frac{\left[K\right(x+x_0)-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}{(1-\frac{x}{r})}^{-2}]}{m}$
$a=\frac{\left[K\right(x+x_0)-\frac{1}{4\pi {ϵ}_0}\frac{q^2}{r^2}(1+\frac{2x}{r})]}{m}$
$a=\frac{[K-\frac{q^2}{2\pi {ϵ}_0{r}^3}]}{m}x$
$\Rightarrow T=2\pi \sqrt{\frac{m}{[K-(\frac{q^2}{2\pi {ϵ}_0{r}^3})}}$
and
for real values of ${}^{\prime }{T}^{\prime }$
$K>\frac{q^2}{2\pi {ϵ}_0{r}^3}$