Oscillations

JEE Main level, single correct, medium

41 of 194

Question 41 of 194

A block of mass 1kg1kg attached to a spring is made to oscillate with an initial amplitude of 12cm12cm. After 22 minutes the amplitude decreases to 6cm6cm. Determine the value of the damping constant for this motion. (take ln⁡2=0.693\ln 2=0.693 )

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Answer: (A) 1.16×10−2kgs−11.16\times {10}^{-2}kg{s}^{-1}

The amplitude of damped oscillation with time tt and damping constant bb is given by,
A=A0e−bt/2mA={A}_{0}{e}^{-bt/2m}
⇒ln⁡(A0A)=bt2m\Rightarrow \ln \left(\dfrac{{A}_{0}}{A}\right)=\dfrac{bt}{2m}
⇒ln⁡2=b2m×120\Rightarrow \ln 2=\dfrac{b}{2m}\times 120; here t=2min⁡=120st=2\min =120s
⇒0.693×2×1120=b\Rightarrow \dfrac{0.693\times 2\times 1}{120}=b
⇒b=1.16×10−2kgs−1\Rightarrow b=1.16\times {10}^{-2}kg{s}^{-1}.

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