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Physics - SHM Question with Solution | TestHub

PhysicsSHMCalculation of time periodMedium2 minQB

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A particle of mass 1 kg is moving where potential energy varies with displacement $U = 10 (1 - cos 2 x)$ , then the time period of small oscillation about origin will be

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Answer:

Solution:

$U = 10 (1 - cos 2 x)$ , so $F = - \frac{d U}{d x} = - 20 sin 2 x$

For small $x$ : $F \approx - 40 x$ , giving $k_{e ff} = 40$ N/m

$T = 2 π \frac{m}{k} = 2 π \frac{1}{40} = π \frac{1}{10}$

Answer: (D)

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

⏱ 2mℹ️ Source: QB

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