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 is connected to an elastic spring of spring constant and oscillates in a damping medium of damping constant . 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
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
Given,
Amplitudeofdampedoscillation = A = {A}_{o}e^{- \dfrac{bt}{2m}}⇒ \dfrac{{A}_{o}}{\sqrt{2}} = {A}_{o}e^{- \dfrac{bt}{2m}}\sqrt{2} = e^{\dfrac{bt}{2m}}\ln \left(\sqrt{2}\right) = \dfrac{bt}{2m} = \dfrac{{10}^{- 2}}{2 \times 0.1}t =\dfrac{0.1t}{2}\dfrac{t}{2} =10\ln \left(\sqrt{2}\right) = 10 \times \dfrac{1}{2} \times 0.693t = 10 \times 0.693 = 6.93t \approx 7s$.