Oscillations

JEE Main level, numerical answer, medium

14 of 194

Question 14 of 194

An object of mass 0.2 kg0.2 \mathrm{~kg} executes simple harmonic motion along x\mathrm{x} axis with frequency of (25π)Hz\left(\frac{25}{\pi}\right) \mathrm{Hz}. At the position x=0.04 mx=0.04 \mathrm{~m} the object has kinetic energy 0.5 J0.5 \mathrm{~J} and potential energy 0.4 J0.4 \mathrm{~J}. The amplitude of oscillation is _____cm\mathrm{cm}.

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Answer: 6

 Total energy = K.E. + P.E.  at x=0.04 m, T.E. =0.5+0.4=0.9 J T.E =1 mω2 A2=0.9=12×0.2(2π×25π)2×A2=0.9⇒A=0.06 m A=6 cm\begin{aligned} & \text { Total energy }=\text { K.E. }+ \text { P.E. } \\ & \text { at } x=0.04 \mathrm{~m}, \text { T.E. }=0.5+0.4=0.9 \mathrm{~J} \\ & \begin{aligned} \text { T.E } & =1 \mathrm{~m} \omega^2 \mathrm{~A}^2=0.9 \\ & =\frac{1}{2} \times 0.2\left(2 \pi \times \frac{25}{\pi}\right)^2 \times \mathrm{A}^2=0.9 \\ \Rightarrow \mathrm{A} & =0.06 \mathrm{~m} \\ \mathrm{~A} & =6 \mathrm{~cm}\end{aligned}\end{aligned}

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