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Two simple harmonic motions,\( y _{1}= a \sin \omega t \) and\( y _{2}=2 a \sin \left(\omega t +\frac{2 \pi}{3}\right) \) aresuperimposed on a particle of mass \( m \). The maximum kinetic energy of the particle is:1. \( \frac{1}{2} m \omega^{2} a^{2} \)2. \( \frac{5 m \omega^{2} a ^{2}}{4} \)3. \( \frac{3}{2} m \omega^{2} a^{2} \)4. Zero |
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Answer» Correct option is (a) \(\frac12\)mω2a2 y1 = a sin ωt y2 = 2a sin(ωt + 2π/3) then Resultant amplitude = \(\sqrt{a^2+(2a)^2+2\times2a\times a cos 120^\circ}\) = \(\sqrt{a^2+4a^2-4a^2} \) = a Maximum kinetic energy E = \(\frac12\)mω2a2 |
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