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PhysicsSHMCalculation of time periodEasy2 minPYQ_2020

PhysicsEasysingle choice

The magnitude of the maximum acceleration is $π$ times that of the maximum velocity of a simple harmonic oscillator. The time period of the oscillator in second is

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Answer:

Solution:

Maximum acceleration = Maximum velocity $\times π$

i.e., $ω^{2} A = π ω A$

where $A$ is amplitude and $ω$ is angular velocity.

$\Rightarrow$ $ω = π$

$\Rightarrow$ $\frac{2 π}{T} = π$

$\Rightarrow$ $T = 2 s$

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

⏱ 2mℹ️ Source: PYQ_2020

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