Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693)
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500g, Decay constant = 20g/s then how much time is required for the amplitude of the system to drop to half of its initial value? (ln 2 = 0.693)
Option 1 - <p>34.65 s</p>
Option 2 - <p>15.01 s</p>
Option 3 - <p>17.32 s</p>
Option 4 - <p>0.034 s</p>
2 Views|Posted 5 months ago
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1 Answer
A
Answered by
5 months ago
Correct Option - 1
Detailed Solution:
As we know that for damping Oscillation
A = A? e? / (²? ) ⇒ t? /? = ln (2)2m/b = (0.693 * 2 * 500)/20 = 34.65 s
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Physics Oscillations 2025
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