Oscillations

JEE Main level, single correct, medium

174 of 194

Question 174 of 194

A block of mass 0.1kg0.1kg is connected to an elastic spring of spring constant 640Nm−1640N{m}^{-1} and oscillates in a damping medium of damping constant 10−2kgs−1{10}^{-2}kg{s}^{-1} . The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to-

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Answer: (D) 7s7s

Given,
E=Eo2E=\dfrac{{E}_{o}}{2}
⇒A=Ao2\Rightarrow A=\dfrac{{A}_{o}}{\sqrt{2}}
Amplitudeofdampedoscillation=A=Aoe−bt2mAmplitudeofdampedoscillation=A={A}_{o}{e}^{-\dfrac{bt}{2m}}

⇒Ao2=Aoe−bt2m\Rightarrow \dfrac{{A}_{o}}{\sqrt{2}}={A}_{o}{e}^{-\dfrac{bt}{2m}}

2=ebt2m\sqrt{2}={e}^{\dfrac{bt}{2m}}

ln⁡(2)=bt2m=10−22×0.1t=ln⁡\left(\sqrt{2}\right)=\dfrac{bt}{2m}=\dfrac{{10}^{-2}}{2\times 0.1}t=0.1t2\dfrac{0.1t}{2}

t2=\dfrac{t}{2}=10ln⁡⁡(2)=10×12×0.69310\ln ⁡\left(\sqrt{2}\right)=10\times \dfrac{1}{2}\times 0.693

t=10×0.693=6.93t=10\times 0.693=6.93

t≈7st\approx 7s.

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