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JEE Main 20179 AprilOnline
Exam
JEE Main
Year
2017
Date
9 April
Mode
Online
Subject
Physics
Chapter
Oscillations

JEE Main 2017 · 9 April · Online · Physics · Oscillations

A block of mass 0 . 1kg is connected to an elastic spring of spring...

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-

Options

  1. A2s2s
  2. B5s5s
  3. C3s3s
  4. D7s7s

Correct answer

D. 7s7s

Step-by-step solution

Given,

E=Eo2E = \dfrac{Eo}{2}
⇒A=Ao2⇒ A = \dfrac{Ao}{2}
Amplitudeofdampedoscillation = A = {A}_{o}e^{- \dfrac{bt}{2m}}<br/><br/><br /><br />⇒ \dfrac{{A}_{o}}{\sqrt{2}} = {A}_{o}e^{- \dfrac{bt}{2m}} <br/><br/>~<br /><br />\sqrt{2} = e^{\dfrac{bt}{2m}}<br/><br/><br /><br />\ln \left(\sqrt{2}\right) = \dfrac{bt}{2m} = \dfrac{{10}^{- 2}}{2 \times 0.1}t =\dfrac{0.1t}{2}<br/><br/><br /><br />\dfrac{t}{2} =10\ln \left(\sqrt{2}\right) = 10 \times \dfrac{1}{2} \times 0.693<br/><br/><br /><br />t = 10 \times 0.693 = 6.93<br/><br/><br /><br />t \approx 7s$.

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