Question

A block having mass m and charge q is connected by 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 released from rest when spring is unstretched (at x=0). Then -

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

Answer 1

Answer: (A), (B), (C)

Maximum stretch in the spring is $\frac{2qE}{k}$

In equilibrium position, stretch in the spring is $\frac{qE}{k}$

Amplitude of oscillation of block is $\frac{qE}{k}$

Answer 2

The block is subjected to two horizontal forces: the electric force $F_e = qE$ and the spring force $F_s = -kx$.

The equilibrium position $x_{eq}$ is where the net force is zero: $qE - kx_{eq} = 0$ $x_{eq} = \frac{qE}{k}$.

The equation of motion can be rewritten in terms of the displacement from the equilibrium position, $x' = x - x_{eq} = x - \frac{qE}{k}$. The amplitude of oscillation is $A = \frac{qE}{k}$.

Using energy conservation, the maximum stretch in the spring is $x_{max} = \frac{2qE}{k}$.