A particle moves such that its acceleration is given by a- x-2- Here - is a positive constant and x is the position from origin- Time period of oscillations is

The correct option is B 2π√(1/β) Acceleration a = ω^2 x Comparing nbsp; ω^2 = β nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; n ...

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Standard XII

Physics

Velocity Displacement Relationship

A particle mo...

Question

A particle moves such that its acceleration is given by $a = - β (x - 2)$ . Here $β$ is a positive constant and $x$ is the position from origin. Time period of oscillations is

2 π \sqrt{β}

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2 π \sqrt{\frac{1}{β}}

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2 π \sqrt{β + 2}

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2 π \sqrt{\frac{1}{β + 2}}

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Solution

The correct option is B $2 π \sqrt{\frac{1}{β}}$
Acceleration $a = - ω^{2} x$
Comparing $ω^{2} = β$
$w = \sqrt{β}$
$T = \frac{2 π}{w} = \frac{2 π}{\sqrt{β}}$

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A particle moves such that its acceleration is given by a=−β(x−2). Here β is a positive constant and x is the position from origin. Time period of oscillations is

The correct option is B 2π√1βAcceleration a=−ω2xComparing ω2=β w=√β T=2πw=2π√β

A particle moves such that its acceleration is given by $a = - β (x - 2)$ . Here $β$ is a positive constant and $x$ is the position from origin. Time period of oscillations is

The correct option is B $2 π \sqrt{\frac{1}{β}}$
Acceleration $a = - ω^{2} x$
Comparing $ω^{2} = β$
$w = \sqrt{β}$
$T = \frac{2 π}{w} = \frac{2 π}{\sqrt{β}}$