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)

A, B, C

Answer 2

The block is subjected to a constant electric force $F_E = qE$ and a spring force $F_s = -kx$. The net force is $F_{net} = qE - kx$.

The equilibrium position $x_{eq}$ is where $F_{net} = 0$: $qE - kx_{eq} = 0 \implies x_{eq} = \frac{qE}{k}$

The equation of motion is $m\frac{d^2x}{dt^2} = qE - kx$. Let $x' = x - x_{eq}$. Then $m\frac{d^2x'}{dt^2} + kx' = 0$. The angular frequency is $\omega = \sqrt{\frac{k}{m}}$.

Initial conditions: $x(0) = 0$ and $v(0) = 0$. Thus, $x'(0) = -\frac{qE}{k}$.

The general solution is $x'(t) = A \cos(\omega t + \phi)$.

$x'(0) = A \cos(\phi) = -\frac{qE}{k}$. $v'(0) = -A\omega \sin(\phi) = 0 \implies \phi = \pi$.

Therefore, $A = \frac{qE}{k}$.

The position of the block is $x(t) = \frac{qE}{k} + \frac{qE}{k} \cos(\omega t + \pi) = \frac{qE}{k} - \frac{qE}{k} \cos(\omega t)$.

$x_{max} = \frac{2qE}{k}$.