JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Elasticity & SHM

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A particle of mass 1 mg and charge q is lying at the mid-point of two stationary particles kept at a distance '2 m' when each is carrying same change 'q'. If the free charged particle is displaced from its equilibrium position through distance '

x

' (

x ≪ 1 m

). The particle executes SHM. Its angular frequency of oscillation will be ____

\times 1 0^{5} rad / s

q^{2} = 10 C^{2}

Answer: 6000

Solution

Each fixed charge is

a = 1

m from the midpoint. When the free charge is pushed a distance

x

along the line towards one of them, the two Coulomb forces no longer cancel:

F = \frac{k q ^{2}}{( a - x ) ^{2}} - \frac{k q ^{2}}{( a + x ) ^{2}} \approx \frac{4 k q ^{2}}{a ^{3}} x

and it points back towards the midpoint, so the motion is simple harmonic with

ω = \frac{4 k q ^{2}}{m a ^{3}} = \frac{4 \times 9 \times 1 0 ^{9} \times 10}{1 0 ^{- 6} \times 1} = 6 \times 1 0^{8} rad/s

In the units asked for this is

6000 \times 1 0^{5}

rad/s.

JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Integer answer

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JEE Main · 2021 · 25 Jul 2021 · Shift 1 · Elasticity & SHM

Solution

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