Verified Solution • 2025–26 Session moderate

QJEEM21Ph896c: JEE Main 2021 Physics - Class 11 JEEM21Ph896c Solution: An object of mass 0.5 kg is executing simple harmonic motion.…

An object of mass 0.5 kg is executing simple harmonic motion. It amplitude is 5 cm and time period (T) is 0.2 s. What will be the potential energy of the object at an instant t=T4st = {T \over 4}s starting from mean position. Assume that the initial phase of the oscillation is zero.

Multiple Choice Options

  • A 0.62 J
  • B 6.2 ×\times 10 −- 3 J
  • C 1.2 ×\times 103 J
  • D 6.2 ×\times 103 J

Step-by-Step Verified Solution

T=2πmkT = 2\pi \sqrt {{m \over k}}

0.2=2π0.5k0.2 = 2\pi \sqrt {{{0.5} \over k}}

k = 50 π\pi 2

≈\approx 500

x = A sin ( ω\omega t + ϕ\phi )

= 5 cm sin (ωT4+0)\left( {{{\omega T} \over 4} + 0} \right)

= 5 cm sin (π2)\left( {{\pi \over 2}} \right)

= 5 cm

PE=12kx2PE = {1 \over 2}k{x^2}

=12(500)(5100)2= {1 \over 2}(500){\left( {{5 \over {100}}} \right)^2}

= 0.6255

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