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A particle is executing simple harmonic motion along a straight line PQ. At three points A, B and C on the line PQ, lying on one side of the mean position, the velocities of the particles are 8ms^(-1), 7ms^(-1) " and "4ms^(-1), respectively. If AB = BC = 1m, the velocity of the particle at mean position is |
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Answer» `9ms^(-1)` velocity of particle A, `v_(A)=8ms^(-1)` velocity of particle B, `v_(B)=7ms^(-1)` velocity of particle C, `v_(C)=4 ms^(-1)` velocity of a particle in harmonic motion (SHM), `RARR ""V=omegasqrt(a^(2)-x^(2))` Square on the both sides, we get `""v^(2)=a^(2)omega^(2)-x^(2)omega^(2)` For particle A, `""v_(A)^(2)=a^(2)omega^(2)-x^(2)omega^(2)` `rArr ""64=a^(2)omega^(2)-x^(2)omega^(2) "...(i)"` For particle B, `""v_(B)^(2)=a^(2)omega^(2)-(x+1)^(2)omega^(2)` `rArr ""49=a^(2)omega^(2)-(x+1)^(2)omega^(2) "...(ii)"` For particle C, `""v_(C)^(2)=a^(2)omega^(2)-(x+2)^(2)omega^(2)` `rArr ""16=a^(2)omega^(2)-(x+2)^(2)omega^(2) "...(iii)"` By solving Eqs. (i), (ii) and (iii), we get `""x=1/3, a^(2)omega^(2)=65 ""(because v=aomega)` HENCE, velocity of particle at the MEAN position, `v=aomega=sqrt(65)ms^(-1)`. |
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