JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Elasticity & SHM

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

\frac{T}{x}

. The value of

x

is _____.

Answer: 8

Solution

With

y = A sin ω t

, the potential energy is

\frac{1}{2} k y^{2}

and the kinetic energy is

\frac{1}{2} k (A^{2} - y^{2})

. They are equal when

A^{2} - y^{2} = y^{2}

, that is

y = \frac{A}{2}

. Then

sin ω t = \frac{1}{2}

, so

ω t = \frac{π}{4}

and

t = \frac{π}{4 ω}

. Since

T = \frac{2 π}{ω}

t = \frac{T}{8} .

Thus

x = 8

JEE Main · 2021 · 31 Aug 2021 · Shift 1 · Numerical answer

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