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A particle executing simple harmonic motion has an amplitude of 6 cm . Its acceleration at a distance of 2 cm from the mean position is `8 cm/s^(2)` The maximum speed of the particle isA. `8 cms^(-1)`B. `12 cms^(-1)`C. `16 cms^(-1)`D. `24 cms^(-1)` |
Answer» Correct Answer - B `a=omega^(2)yimpliesomega=sqrt((a)/(y))=sqrt((8)/(2))=2"rads"^(-1)` Now `v_("max")=Aomega=6xx 2=12 cms^(-1)` |
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