JEE Mains · Physics · STD 11 - 13. oscillations
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass \(=500\, g\), Decay constant \(=20 \,g / s\) then ...... \(s\) time is required for the amplitude of the system to drop to half of its initial value ? \((\ln 2=0.693)\)
- A \(34.65\)
- B \(17.32\)
- C \(0.034\)
- D \(15.01\)
Answer & Solution
Correct Answer
(A) \(34.65\)
Step-by-step Solution
Detailed explanation
\(A = A _{0} e ^{-\gamma t }= A _{0} e ^{-\frac{ bt }{2 m }}\) \(\frac{ A _{0}}{2}= A _{0} e ^{-\frac{ bt }{2 m }}\) \(\frac{ bt }{2 m }=\ln 2\) \(t =\frac{2 m }{ b } \ln 2=\frac{2 \times 500 \times 0.693}{20}\) \(t =34.65\, second.\)
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