A wooden block performs SHM on a frictionless | vedclass

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(A) The frequency of oscillation for a block-spring system is given by $v_0 = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.This frequency depends only on the s

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of the same frequency and with shifted mean position.

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(A) The frequency of oscillation for a block-spring system is given by $v_0 = \frac{1}{2 \pi} \sqrt{\frac{k}{m}}$.This frequency depends only on the spring constant $k$ and the mass $m$ of the block,and it is independent of any constant external force.When a uniform electric field $\vec{E}$ is applied,a constant electrostatic force $F_e = QE$ acts on the block.This force shifts the equilibrium (mean) position of the block to a new position where the spring force balances the electric force,i.e.,$kx' = QE$,where $x'$ is the displacement from the original mean position.Since the force is constant,it does not affect the restoring force gradient (the spring constant $k$),and therefore the frequency of oscillation remains unchanged.Thus,the block performs $SHM$ with the same frequency but about a new,shifted mean position.Therefore,option $(a)$ is correct.

$A$ wooden block performs $SHM$ on a frictionless surface with frequency $v_0$. The block carries a charge $+Q$ on its surface. If now a uniform electric field $\vec{E}$ is switched-on as shown,then the $SHM$ of the block will be

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