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Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodEasy2 minPYQ_2020

PhysicsEasysingle choice

The maximum velocity of a particle, executing simple harmonic motion with an amplitude $7 mm$ is $4.4 m s^{- 1}$ . The period of oscillation is

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Answer:

Solution:

Maximum velocity $ν_{m} = a ω = a (\frac{2 π}{T})$

$∴$ $T = \frac{2 π a}{ν_{m}} = 2 \times \frac{22}{7} \times \frac{(7 \times 10^{- 3})}{4 \cdot 4}$

$T = 10^{- 2} = 0.01 s$

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

⏱ 2mℹ️ Source: PYQ_2020

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