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Standard XII

Physics

Relative Velocity in 2D

A particle mo...

Question

A particle moves such that its acceleration $^{'} a^{'}$ is given by $a = - b x$ where $x$ is the displacement from equilibrium position and $b$ is constant. The period of oscillation

2 π / b

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2 π / \sqrt{b}

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\sqrt{2 π / b}

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2 \sqrt{π / b}

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Solution

The correct option is A $2 π / \sqrt{b}$
$a = - ω^{2} x = - b x$
Thus, $ω = \sqrt{b}$
Thus, $T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{b}}$

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A particle moves such that its acceleration ′a′ is given by a=−bx where x is the displacement from equilibrium position and b is constant. The period of oscillation

The correct option is A 2π/√ba=−ω2x=−bxThus, ω=√bThus, T=2πω=2π√b

A particle moves such that its acceleration $^{'} a^{'}$ is given by $a = - b x$ where $x$ is the displacement from equilibrium position and $b$ is constant. The period of oscillation

The correct option is A $2 π / \sqrt{b}$
$a = - ω^{2} x = - b x$
Thus, $ω = \sqrt{b}$
Thus, $T = \frac{2 π}{ω} = \frac{2 π}{\sqrt{b}}$