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Physics - SHM Question with Solution | TestHub

PhysicsSHMKinematicsEasy2 minQB

PhysicsEasysingle choice

A particle performs SHM with an amplitude of $5 m$ . If at $x = 4 m$ , the magnitude of velocity and acceleration are equal, then the time period of SHM (in seconds) is

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Answer:

Solution:

$v = ω \sqrt{5^{2} - 4^{2}} = 3 ω$

$a = ω^{2} \times 4$

$|a| = |v|$

$4 ω^{2} = 3 ω$

$ω = \frac{3}{4} = \frac{2 π}{T}$

$T = \frac{8 π}{3} s$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Kinematics

⏱ 2mℹ️ Source: QB

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