A block of mass 1 kg attached to a spring is made to oscillate with an initial amplitude of 12 cm. After 2 minutes the amplitude decreases to 6 cm. Determine the value of the damping constant for this motion . (take ln 2 = 0.693)
Solution
$A = {A_o}{e^{{{ - b} \over {2m}}t}}$<br><br>$\Rightarrow$ $6 = 12{e^{{{ - b} \over {2 \times 1}} \times 120}}$<br><br>$\Rightarrow$ $6 = 12{e^{ - b \times 60}}$<br><br>$\Rightarrow$ ${1 \over 2} = {e^{ - 60b}}$<br><br>$\Rightarrow$ $\ln (2) = 60b$<br><br>$\Rightarrow$ $b = {{\ln (2)} \over {60}} = 1.16 \times {10^2}$ Kg/s
About this question
Subject: Physics · Chapter: Oscillations · Topic: Simple Harmonic Motion
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