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The following figure shows a block of mass m suspended from a fixed point by means of a vertical spring. The block is oscillating simple harmonically and carries a charge q. There also exists a uniform electric field in the region. Consider four different cases. The electric field is zero in case 1, mg/q downward in case 2, mg/q upward in case 3, and 2mg/q downward in case 4. The speed at mean position is same in all cases. Select the correct alternative(s).

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Time periods of oscillation are equal in case 1 and case 3.

Amplitudes of displacement are same in case 2 and case 3.

The maximum elongation (increment in length from natural length) is maximum in case 4.

Time periods of oscillation are equal in case 2 and case 4.

answer is A, B, C, D.

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Detailed Solution

(1) Only equilibrium position changes. Time period remains same

$T = 2 π \sqrt{m / k}$ .

(2) $\frac{1}{2} {kx}^{2} = \frac{1}{2} {mv}^{2}$

Since v at mean position is same amplitude will be same.

(3) In case 4 equilibrium position is $x_{0} = 3 mg / k$ . So maximum elongation = $x_{0} + A m p l i t u d e .$

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