Verified Solution • 2025–26 Session moderate

QJEEM23Ph9c19: JEE Main 2023 Physics - Class 11 JEEM23Ph9c19 Solution: The variation of kinetic energy (KE) of a particle executing…

The variation of kinetic energy (KE) of a particle executing simple harmonic motion with the displacement (x)(x) starting from mean position to extreme position (A) is given by

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/0ed097fa73bbfdfde7291831a0f08f12a1660a7df1bb3431ca65c622197889d2.webp
  • B https://ik.imagekit.io/novatra/photos/7f6db029968841eb0702e27238e0b0cc08b24148fd54140c8de20baf1ccecfcd.webp
  • C https://ik.imagekit.io/novatra/photos/a23255f33255d482abc49a9b8bf2b24e798a893f821571f8e9ee3fdd711c6cb2.webp
  • D https://ik.imagekit.io/novatra/photos/e74d0298b93539e05b67e03b892f18765526171d02468b2cda95c705189cb5c0.webp

Step-by-Step Verified Solution

For a particle executing SHM

KE=12mω2(A2−x2)\mathrm{KE}=\frac{1}{2} \mathrm{m} \omega^2\left(\mathrm{A}^2-\mathrm{x}^2\right)

When x=0,\mathrm{x}=0, KE is maximum & when x=A,\mathrm{x}=\mathrm{A}, KE is zero and KE vs x graph is parabola.

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