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PhysicsSHMCalculation of time periodEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle moves so that its acceleration $a$ is given by $a = - bx$ , where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

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Answer:

Solution:

The relation between acceleration ( $a$ ) and displacement ( $x$ ) for a body in SHM is

$a = - ω^{2} x$

Given, $a = - bx$

On comparing the two equations, we get

$ω^{2} = b$

$∴$ $ω = \sqrt{b}$

Since, $ω = \frac{2 π}{T}$

$∴$ $\frac{2 π}{T} = \sqrt{b}$

$\Rightarrow T = \frac{2 π}{\sqrt{b}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Calculation of time period

⏱ 2mℹ️ Source: PYQ_2020

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