An object of mass 0.2 kg executes simple harmonic motion along x axis with frequency of ( π 25 ) Hz . At the position x = 0.04 m the object has kinetic energy 0.5 J and potential energy 0.4 J . The amplitude of oscillation is _____ cm .

Total energy = K.E. + P.E. at x = 0.04 m , T.E. = 0.5 + 0.4 = 0.9 J T.E ⇒ A A = 1 m ω 2 A 2 = 0.9 = 2 1 × 0.2 ( 2 π × π 25 ) 2 × A 2 = 0.9 = 0.06 m = 6 cm

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PhysicsSHMEnergyEasy2 minPYQ_2024

PhysicsEasynumerical

An object of mass $0.2 kg$ executes simple harmonic motion along $x$ axis with frequency of $(\frac{25}{π}) Hz$ . At the position $x = 0.04 m$ the object has kinetic energy $0.5 J$ and potential energy $0.4 J$ . The amplitude of oscillation is _____ $cm$ .

Answer:

6.00

Solution:

$Total energy = K.E. + P.E. at x = 0.04 m, T.E. = 0.5 + 0.4 = 0.9 J T.E \Rightarrow A A = 1 m ω^{2} A^{2} = 0.9 = \frac{1}{2} \times 0.2 (2 π \times \frac{25}{π})^{2} \times A^{2} = 0.9 = 0.06 m = 6 cm$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Energy

⏱ 2mℹ️ Source: PYQ_2024

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