Question

A Non-conducting block performs $\text{SHM}$ on a frictionless surface with frequency, ${\nu }_0$. The block carries a charge $+Q$ on its surface. If a uniform electric field $\overrightarrow{E}$ is switched-on as shown, then the $\text{SHM}$ of the block will be

• A. of the same frequency and with shifted mean position
• B. of the same frequency and with the same mean position
• C. of changed frequency and with shifted mean position
• D. of changed frequency and with the same mean position

Answer 1

The correct option is A of the same frequency and with shifted mean position

$\text{Case I}:$ Effect on the mean position

For the spring block system, if block is charged by $Q$ and placed in an electric field of strength $E$, then there will be electric force on the block given by,$$F_{electric}=QE....\left(1\right)$$


Now in order to maintain equilibrium spring will be compressed by $x_0$, which will impose the force on the spring given by,$$F_{spring}=k{x}_0....\left(2\right)$$Where, $k$ is spring constant.

So, for the equilibrium condition,
$F_{electric}=F_{spring}$

$\Rightarrow QE=k{x}_0$

$\therefore x_0=\frac{QE}{k}$

So, the mean position is shifted from natural length by $\left(\frac{QE}{k}\right)$ distance.

$\text{Case II}:$ Effect on the frequency:

Frequency of oscillation of block, $${\nu }_0=2\pi \sqrt{\frac{k}{m}}$$ So, the frquency does not depend on constant external force due to electric field, $QE$.
Hence, the frequency of oscillation will remain the same.
But due to constant external force, the mean position of the wooden block alone gets shifted .

Hence, option (a) is the correct alternative.