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Example 14.10 For the damped oscillator the mass m of the block is 200 g, k = 90 N $m^{- 1}$ and the damping constant b is 40 g $s^{- 1}$ . Calculate (a) the period of oscillation, (b) time taken for its amplitude of vibrations to drop to half of its initial value and (c) the time taken for its mechanical energy to drop to half its initial value.

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Solution

Dear Student,
$m = 0.2 k g, K = 90 N / m, b = 0.04 k g / s ω = \sqrt{\frac{K}{m} - {(\frac{b}{2 m})}^{2}} = \sqrt{\frac{90}{0.2} - {(\frac{0.04}{0.4})}^{2}} = 21.21 r a d / s T = \frac{2 π}{21.21} = 0.3 s e c . f o l l o w t h e e q n A = A_{0} e^{(\frac{- b t}{2 m})} A = \frac{A_{0}}{2} \frac{A_{0}}{2} = A_{0} e^{(\frac{- b t}{2 m})} (\frac{- b t}{2 m}) = - \ln 2 t = \frac{\ln 2 * 2 m}{b} = \frac{0.693 * 2 * 0.2}{0.04} = 6.93 s e c i n i t i a l e n e r g y o f t h e s y s t e m E = \frac{K {A_{0}}^{2}}{2} l e t a t A t h e s y s t e m e n e r g y b e c o m e s h a l f \frac{K A^{2}}{2} = \frac{K {A_{0}}^{2}}{4} A = \frac{A_{0}}{\sqrt{2}} t i m e t a k e n t o r e a c h t i s v a l u e . \frac{A_{0}}{\sqrt{2}} = A_{0} e^{(\frac{- b t}{2 m})} \frac{- b t}{2 m} = - \ln \sqrt{2} t = \frac{\ln \sqrt{2} * 2 m}{b} = \frac{\ln \sqrt{2} * 2 * 0.2}{0.04} = 3.5 s e c . r e g a r d s$

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Similar questions

Q. For the damped oscillator shown in Fig, the mass of the block is

200 g, k = 80 N m^{- 1}

and the damping constant

b

40 g s^{- 1}

Calculate.
(a) The period of oscillation,
(b) Time period for its amplitude of vibrations to drop to half of its initial value
(c) The time for the mechanical energy to drop to half initial value.

For the damped oscillator shown, the mass m of the block is 200 g, k = 90 N m $^{- 1}$ and the damping constant b is 40 g s $^{- 1}$ . What is the period of oscillation, time taken for its amplitude of vibrations to drop to half of its initial value.