A block of mass 1 kg attached to a spring is made to oscillate with initial amplitude of 12cm. After 2 minutes the amplitude decreases to 6cm. Determine the value of the damping constant for this motion. (take ln 2 = 0.693)
A block of mass 1 kg attached to a spring is made to oscillate with initial amplitude of 12cm. After 2 minutes the amplitude decreases to 6cm. Determine the value of the damping constant for this motion. (take ln 2 = 0.693)
Option 1 -
3.3 * 10⁻² kg s⁻¹
Option 2 -
1.16 * 10⁻² kg s⁻¹
Option 3 -
0.69 * 10⁻² kg s⁻¹
Option 4 -
5.7 * 10⁻³ kg s⁻¹
Asked by Shiksha User
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1 Answer
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Correct Option - 4
Detailed Solution:Given the decay equation A = A? e^ (-bt/m):
-bt/m = ln (A/A? )
Solving for b:
b = (-m/t? ) * ln (A/A? ) = (-1 / (2 * 60) * ln (6/12) = 5.775 * 10? ³ kg/s
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