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A particle performs simple harmonic motion wit amplitude A. its speed is double at the instant when it is at distance `(A)/(3)` from equilibrium position. The new amplitude of the motion isA. `sqrt(11)A`B. `(sqrt(22)A)/(3)`C. `(sqrt(33)A)/(2)`D. none of these |
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Answer» Correct Answer - B `V=omegasqrt(A^(2)-((A)/(3))^(2))` `V=sqrt((8A^(2)omega)/(9))=(2sqrt(2))/(3)Aomega`. `V_(n ew)=2V=(4sqrt(2))/(3)(A)omega` so the new amplitude is given by `V_(n ew)=omegasqrt((A_(n ew))^(2)-x^(2))` `(4sqrt(2))/(3)Aomega=omegasqrt((A_(n ew))^(2)-((A)/(3))^(2))` `(32)/(9)A^(2)=(A_(n ew))^(2)-(A^(2))/(9)` `A_(n ew)2=(33A^(2))/(9)` `A_(n ew)=(sqrt(33)A)/(3)` |
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