ExamBro
ExamBro
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 ? \((\ln 2=0.693)\)

  1. A \(34.65\)
  2. B \(17.32\)
  3. C \(0.034\)
  4. D \(15.01\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(34.65\)

Step-by-step Solution

Detailed explanation

\(A = A _{0} e ^{-\gamma t }= A _{0} e ^{-\frac{ bt }{2 m }}\) \(\frac{ A _{0}}{2}= A _{0} e ^{-\frac{ bt }{2 m }}\) \(\frac{ bt }{2 m }=\ln 2\) \(t =\frac{2 m }{ b } \ln 2=\frac{2 \times 500 \times 0.693}{20}\) \(t =34.65\, second.\)