InterviewSolution
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InterviewSolution
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A particle executes simple harmonic motion with an amplitude of `4cm` At the mean position the velocity of the particle is `10`cm/s. distance of the particle from the mean position when its speed `5` cm/s isA. `sqrt(3)` cmB. `sqrt(5)` cmC. `2sqrt(3)` cmD. `2sqrt(5)` cm |
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Answer» Correct Answer - C `v_("max")=aomegaimpliesomega=(v_("max"))/(a)=(10)/(4)` Now, `v=omegasqrt(a^(2)-y^(2))` `implies v^(2)=omega^(2)(a^(2)-y^(2))` `implies y^(2)=a^(2)-(v^(2))/(omega^(2))` `implies y=sqrt(a^(2)-(v^(2))/(omega^(2)))=sqrt(4^(2)-(5^(2))/((10//4)^(2)))=2sqrt(3)` cm |
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