JEE MAIN - Physics (2021 - 31st August Morning Shift - No. 20)
A particle of mass 1 kg is hanging from a spring of force constant 100 Nm$$-$$1. The mass is pulled slightly downward and released so that it executes free simple harmonic motion with time period T. The time when the kinetic energy and potential energy of the system will become equal, is $${T \over x}$$. The value of x is _____________.
Answer
8
Explanation
_31st_August_Morning_Shift_en_20_1.png)
KE = PE
$$y = {A \over {\sqrt 2 }} = A\sin \omega t$$
_31st_August_Morning_Shift_en_20_2.png)
$$t = {T \over 8} = {T \over x}$$
x = 8
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