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PhysicsSHMMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle executing SHM has a maximum speed of $0.5 m s^{- 1}$ and a maximum acceleration of $1.0 m s^{- 2}$ . The angular frequency of oscillation is

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Answer:

Solution:

In SHM,

Maximum velocity $v_{m a x} = ω A$

Maximum acceleration $a_{m a x} = A ω^{2}$

$\Rightarrow ω = \frac{A ω^{2}}{A ω} = \frac{a_{m a x}}{v_{m a x}}$

$v_{m a x} = 0.5 m s^{- 1}$

$a_{m a x} = 1.0 m s^{- 1}$

$ω = \frac{1}{0.5} = \frac{10}{5} = 2 rad s^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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