Verified Solution • 2025–26 Session moderate

QJEEM23Phda65: JEE Main 2023 Physics - Class 11 JEEM23Phda65 Solution: The maximum potential energy of a block executing simple…

The maximum potential energy of a block executing simple harmonic motion is 25 J25 \mathrm{~J} . A is amplitude of oscillation. At A/2\mathrm{A / 2} , the kinetic energy of the block is

Multiple Choice Options

  • A 9.75 J
  • B 37.5 J
  • C 18.75 J
  • D 12.5 J

Step-by-Step Verified Solution

Umax⁡=12 mω2 A2=25 J\mathrm{U}_{\max }=\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}=25 \mathrm{~J}

KE\mathrm{KE} at A2=12mv12=12mω2(A2−A24)\frac{\mathrm{A}}{2}=\frac{1}{2} m v_{1}^{2}=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)

=12 mω23 A24=34(12 mω2 A2)=\frac{1}{2} \mathrm{~m} \omega^{2} \frac{3 \mathrm{~A}^{2}}{4}=\frac{3}{4}\left(\frac{1}{2} \mathrm{~m} \omega^{2} \mathrm{~A}^{2}\right)

KE=34×25=18.75 J\mathrm{KE}=\frac{3}{4} \times 25=18.75 \mathrm{~J}

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