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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

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Amplitude of oscillation of block is $\frac{q E}{k}$

Amplitude of oscillation is $\frac{2 q E}{k}$

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

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

answer is A, B, C.

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Detailed Solution

$q E x_{\max} = \frac{1}{2} k x_{\max}^{2} \Rightarrow x_{\max} = \frac{2 q E}{k}$

At Equilibrium, we have

$q E = k x_{eq} \Rightarrow x_{eq} = \frac{q E}{k}$

Amplitude of oscillation is equal to the displacement of block from the mean position, So, amplitude is

$A = \frac{2 q E}{k} - \frac{q E}{k} = \frac{q E}{k}$

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