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JEE PHYSICS MCQ: OSCILLATIONS

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A block whose mass is 650 g650 \mathrm{~g} is fastened to a spring whose spring constant is 65 N/m65 \mathrm{~N} / \mathrm{m}. The block is pulled a distance x=11 cmx=11 \mathrm{~cm} from its equilibrium position at x=0x=0 on a frictionless surface and released from rest at t=0t=0. The maximum velocity of the vibrating block is
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Correct option: A
1.1 m/s1.1 \mathrm{~m} / \mathrm{s}
Option A (1.1 m/s1.1 \mathrm{~m} / \mathrm{s}) is correct because the maximum velocity is calculated by equating the potential energy stored in the spring to the kinetic energy of the block.

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