Question

Two charges of charge $+1\text{C}$ are connected by a spring of spring constant $100\text{N/m}$ and length $1\text{m}$. A small displacement in charge $2$ results into oscillation of the system. Calculate the time period of oscillation.
(mass of charge is $1\text{gm})$

• A. $1.5\mu \text{s}$
• B. $2.2\mu \text{s}$
• C. $3.1\mu \text{s}$
• D. $1.0\mu \text{s}$

Answer 1

The correct option is A $1.5\mu \text{s}$


After doing a small displacement $x$ in charge $2$ as shown in figure,


Here, net restoring force is,

$F=kx-F_e=kx-\frac{1\times 1}{4\pi {ϵ}_0{(l_0+x)}^2}$

Differentiating both side w.r.t. $x$,

$F^{\prime }\left(x\right)=k-\frac{1\times (-2)}{4\pi {ϵ}_0{(l_0+x)}^3}$

$\therefore F^{\prime }\left(0\right)=k+\frac{1}{2\pi {ϵ}_0{l}_0^3}$
$=100+\frac{1}{2\pi {ϵ}_0(1{)}^3}$

$=100+\frac{1}{2\pi {ϵ}_0}=100+(0.017\times {10}^{12})$

Time period of the system will be,

$\therefore T=2\pi \sqrt{\frac{m}{F^{\prime }\left(0\right)}}$

$T=2\pi \sqrt{\frac{{10}^{-3}}{100+(0.017\times {10}^{12})}}$

$\therefore T=1.5\times {10}^{-6}\text{s}=1.5\mu \text{s}$

Hence, option $\left(\text{A}\right)$ is the right choice.

Why this question ? This question is based on Taylor's theorem and according to this, $T=2\pi \sqrt{\frac{m}{F^{\prime }\left(0\right)}}$