Verified Solution • 2025–26 Session moderate

QJEEM17Ph2314: JEE Main 2017 Physics - Class 11 JEEM17Ph2314 Solution: A particle is executing simple harmonic motion with a time…

A particle is executing simple harmonic motion with a time period T. At time t = 0, it is at its position of equilibrium. The kinetic energy – time graph of the particle will look like:

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/9a8c02b1cb733100b6b1b7c84d60fc58436a2fb0e706c73df6c6489fdc1c65be.webp
  • B https://ik.imagekit.io/novatra/photos/8f3eaa2aebfdb1359005f0c1b95388d650ebd0c806d66706674cef6a868368bb.webp
  • C https://ik.imagekit.io/novatra/photos/221b50ae01c3cac896c5304a10b7cff659e03a0cf26bfe01f9af07a7fd9dc50d.webp
  • D https://ik.imagekit.io/novatra/photos/b5e104724e51b988a7385b0d81d0c4ee8f836d94b7a073a0b87bfd1ff8caf5ef.webp

Step-by-Step Verified Solution

For a particle executing SHM

At mean position; t = 0, ω\omega t = 0, y = 0, V = Vmax = a ω\omega

∴\therefore K.E = K.Emax = 12mω2a2{1 \over 2}m{\omega ^2}{a^2}

At extreme position : t = T4{T \over 4} , ω\omega t = π2{\pi \over 2} , y = A, V = Vmin = 0

∴\therefore K.E = K.Emin = 0

Hence graph (A) correctly represents kinetic energy - time graph.

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