JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Elasticity & SHM

The potential energy of a particle of mass 4 kg in motion along the x-axis is given by

U = 4 (1 - cos 4 x)

J. The time period of the particle for small oscillation

(sin θ ≃ θ)

(\frac{π}{K})

s. The value of

K

is ____

Answer: 2

Solution

The force on the particle is

F = - \frac{d U}{d x} = - 4 \cdot 4 sin (4 x) = - 16 sin (4 x)

. For small oscillations,

sin θ \approx θ

, so

F \approx - 16 \cdot 4 x = - 64 x

. The acceleration is

a = F / m = - 64 x /4 = - 16 x

. Comparing with SHM equation

a = - ω^{2} x

, we get

ω^{2} = 16

, so

ω = 4

rad/s. The time period is

T = \frac{2 π}{ω} = \frac{2 π}{4} = \frac{π}{2}

s. Comparing with

T = \frac{π}{K}

, we get

K = 2

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer