Oscillations

JEE Main level, single correct, medium

157 of 194

Question 157 of 194

The maximum potential energy of a block executing simple harmonic motion is 25J25J. A is amplitude of oscillation. At A2\dfrac{A}{2}, the kinetic energy of the block is

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Answer: (C) 18.75J18.75J

The maximum potential energy of a block executing simple harmonic motion is given by Umax⁡=12mω2A2=25J{U}_{\max }=\dfrac{1}{2}m{\omega }^{2}{A}^{2}=25JAs, kinetic energy of particle executing SHM is KE=12mω2(A2−x2)KE=\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-{x}^{2}\right).So, kinetic energy of block KEatA2=12mv12=12mω2(A2−A24)KEat\dfrac{A}{2}=\dfrac{1}{2}m{v}_{1}^{2}=\dfrac{1}{2}m{\omega }^{2}\left({A}^{2}-\dfrac{{A}^{2}}{4}\right)
⇒KE=12mω23A24=34(12mω2A2)\Rightarrow KE=\dfrac{1}{2}m{\omega }^{2}\dfrac{3{A}^{2}}{4}=\dfrac{3}{4}\left(\dfrac{1}{2}m{\omega }^{2}{A}^{2}\right)
⇒KE=34×25=18.75J\Rightarrow KE=\dfrac{3}{4}\times 25=18.75J

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