The maximum potential energy of a block executing simple harmonic motion is 25 J. A is amplitude of oscillation. At A/2, the kinetic energy of the block is (a) 9.75 J (b) 37.5 J (c) 18.75 J (d) 12.5 J

Answer: (c). Maximum PE = total energy E = 25 J. At x = A/2: PE = E/4 = 6.25 J. KE = 25 - 6.25 = 18.75 J So the answer is (c) 18.75 J.

The maximum potential energy of a block executing simple harmonic… | JEE Main 2023 (31 Jan, Shift 1)

Q40JEE Main 2023 · 31 JAN · SHIFT 1Easy

Energy in SHM

The maximum potential energy of a block executing simple harmonic motion is $25\text{ J}$ . A is amplitude of oscillation. At $A/2$ , the kinetic energy of the block is

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Answer: (c)

Maximum PE = total energy $E = 25$ J.
At $x = A/2$ : $PE = \frac{E}{4} = 6.25$ J.
$KE = 25 - 6.25 = 18.75$ J
So the answer is (c) 18.75 J.

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