Verified Solution • 2025–26 Session moderate

QJEEM19Ph8152: JEE Main 2019 Physics - Class 11 JEEM19Ph8152 Solution: A particle executes simple harmonic motion with an amplitude of …

A particle executes simple harmonic motion with an amplitude of 5 cm. When the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is -

Multiple Choice Options

  • A 4π3{{4\pi } \over 3}
  • B 38π{3 \over 8}\pi
  • C 73π{7 \over 3}\pi
  • D 8π3{{8\pi } \over 3}

Step-by-Step Verified Solution

v=ωA2−x2  v = \omega \sqrt {{A^2} - {x^2}} \,\,     . . .(1)

a=−ω2xa = - {\omega ^2}x                . . .(2)

∣v∣=∣a∣\left| v \right| = \left| a \right|                    . . .(3)

ωA2−x2=ω2x\omega \sqrt {{A^2} - {x^2}} = {\omega ^2}x

A2−x2=ω2x2{A^2} - {x^2} = {\omega ^2}{x^2}

52−42=ω2(42){5^2} - {4^2} = {\omega ^2}\left( {{4^2}} \right)

⇒   3=ω×4\Rightarrow \,\,\,3 = \omega \times 4

T=2π/ωT = 2\pi /\omega

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