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Simple Harmonic Motion question

2022 · Shift 2 · Q37
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  5. /2022 · Shift 2 · Q37

Simple Harmonic Motion question

2022 · Shift 2 · Q37

JEE AdvancedPhysicsSimple Harmonic MotionNumerical+3 / −1
A particle of mass 1 kg1 \mathrm{~kg}1 kg is subjected to a force which depends on the position as F⃗=−k(xı^+yȷ^)kg ms−2\vec{F}=-k(x \hat{\imath}+y \hat{\jmath}) \mathrm{kg}\, \mathrm{m} \mathrm{s}^{-2}F=−k(x^+y^​)kgms−2 with k=1 kg s−2k=1 \mathrm{~kg} \mathrm{~s}^{-2}k=1 kg s−2. At time t=0t=0t=0, the particle's position r⃗=(12ı^+2ȷ^)m\vec{r}=\left(\frac{1}{\sqrt{2}} \hat{\imath}+\sqrt{2} \hat{\jmath}\right) mr=(2​1​^+2​^​)m and its velocity v⃗=(−2ı^+2ȷ^+2πk^)ms−1\vec{v}=\left(-\sqrt{2} \hat{\imath}+\sqrt{2} \hat{\jmath}+\frac{2}{\pi} \hat{k}\right) m s^{-1}v=(−2​^+2​^​+π2​k^)ms−1. Let vxv_{x}vx​ and vyv_{y}vy​ denote the xxx and the yyy components of the particle's velocity, respectively. Ignore gravity. When z=0.5 mz=0.5 \mathrm{~m}z=0.5 m, the value of (xvy−yvx)\left(x v_{y}-y v_{x}\right)(xvy​−yvx​) is ‾\underline{\hspace{2cm}}​m2s−1m^{2} s^{-1}m2s−1.
Numerical answer
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