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Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodEasy2 minPYQ_2005

PhysicsEasysingle choice

A particle executing simple harmonic motion of amplitude $5 cm$ has maximum speed of $31.4 cm / s$ . The frequency of its oscillation is:NEET 2005

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Answer:

Solution:

Here $a = 5 cm, V_{m a x} = \frac{31.4 cm}{s}$ $V_{m a x} = ωa = 31.4 = 2 π v \times 5 \Rightarrow 31.4 = 10 \times 3.14 \times V \Rightarrow V = 1 Hz$

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

⏱ 2mℹ️ Source: PYQ_2005

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