Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass $= 500 g$ , damping constant $= 20 g/s$ then how much time is required for the amplitude of the system to drop to half of its initial value?
$(l n 2 = 0.693)$

15.01 s

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17.32 s

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0.034 s

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34.65 s

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Solution

The correct option is D $34.65 s$
In damped oscillation, amplitude is given as: $A = A_{0} e^{- \frac{b t}{2 m}}$ From question, $A = \frac{A_{0}}{2}$
$\Rightarrow \frac{1}{2} = e^{- \frac{b t}{2 m}}$

$\Rightarrow \frac{b t}{2 m} = l n 2 = 0.693$

$\Rightarrow t = \frac{2 m}{b} \times 0.693$

$\Rightarrow t = 2 \times \frac{500}{20} \times 0.693$

$∴ t = 50 \times 0.693 = 34.6 s$

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Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass =500 g, damping constant =20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln2=0.693)

The correct option is D 34.65 sIn damped oscillation, amplitude is given as: A=A0e−bt2m From question, A=A02 ⇒12=e−bt2m ⇒bt2m=ln2=0.693 ⇒t=2mb×0.693 ⇒t=2×50020×0.693 ∴t=50×0.693=34.6 s

Amplitude of a mass spring system, which is executing simple harmonic motion decreases with time. If mass $= 500 g$ , damping constant $= 20 g/s$ then how much time is required for the amplitude of the system to drop to half of its initial value?
$(l n 2 = 0.693)$