JEE Mains · Physics · STD 11 - 13. oscillations
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 In \(2=0.693\) )
- A \(0.69 \times 10^{2}\, kg s ^{-1}\)
- B \(3.3 \times 10^{2}\, kg s ^{-1}\)
- C \(5.7 \times 10^{-3}\, kg s ^{-1}\)
- D \(1.16 \times 10^{2} \,kg s ^{-1}\)
Answer & Solution
Correct Answer
(D) \(1.16 \times 10^{2} \,kg s ^{-1}\)
Step-by-step Solution
Detailed explanation
\(A = A _{0} e ^{-\gamma\, t}\) \(\ln 2=\frac{ b }{2 m } \times 120\) \(\frac{0.693 \times 2 \times 1}{120}= b\) \(1.16 \times 10^{-2} \,kg / sec\)
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