Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)
Solution
$A = {A_0}{e^{ - {{bt} \over {2m}}}}$<br><br>${{bt} \over {2m}} = \ln 2 = 0.693$<br><br>$t = {{2m} \over b} \times 0.693$<br><br>$t = 2 \times {{500} \over {20}} \times 0.693$<br><br>$t = 50 \times 0.693 = 34.6$ sec.
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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