Question

A block having mass $m$ and charge $q$ is connected by a spring of force constant $k$. The block lies on a frictionless horizontal track and a uniform electric field $E$ acts on system as shown. The block is relased from rest when the spring is unstretched $(atx=0)$. Identify the wrong statement.

• A. maximum extension in the spring is $\frac{2qE}{k}$
• B. In equilibrium position extension in the spring is $\frac{qE}{k}$
• C. Amplitude of the oscillation of block is $\frac{qE}{k}$
• D. Amplitude of the oscillation is $\frac{2qE}{k}$

Answer 1

The correct option is D Amplitude of the oscillation is $\frac{2qE}{k}$

At maximum extension(Using Work Energy Theorem)
$qE{x}_{max}=\frac{1}{2}k{x}_{max}^2\Rightarrow x_{max}=\frac{2qE}{k}$
At equilibrium, we have,
$qE=k{x}_{eq}\Rightarrow x_{eq}=\frac{qE}{k}$
Also amplitude of oscillation is equal to the displacement of the block from the mean position. So, the amplitude is,
$A=\frac{2qE}{k}-\frac{qE}{k}=\frac{qE}{k}$