AP EAMCET · PHYSICS · Oscillations
A block of mass \(100 \mathrm{~g}\) is connected to an elastic spring of spring constant \(450 \mathrm{Nm}^{-1}\) oscillates vertically. The block-spring system is in viscous surrounding medium with a damping constant \(69.3 \mathrm{~g} \mathrm{~s}^{-1}\). The time in which the amplitude of oscillations drop to half of its initial value.
(take, \(\ln 2=0.693)\)
- A \(6.93 \mathrm{~s}\)
- B \(2 \mathrm{~s}\)
- C \(20 \mathrm{~s}\)
- D \(69.3 \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(2 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
1 ) Amplitude of a damped oscillator varies with time as \[ A=A_{0 .} e^{-\alpha t} \] Here, \(\alpha=b / 2 m, b=\) damping constant and \(m=\) mass of oscillator. Here, given \(A=A_0 / 2\) So, from eq. (i) we have…
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