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If the KE of a particle performing a SHM of amplitude A is `(3)/(4)` of its total energy, then the value of its displacement isA. `x=pm(A)/(2)`B. `x=pm(A)/(4)`C. `x=pm(sqrt(3A))/(2)`D. `x=pm(A)/(sqrt(2))` |
Answer» Correct Answer - A `K.E.=(3)/(4)T.E.` `(1)/(2)momega^(2)(A^(2)-x^(2))=(3)/(4)xx(1)/(2)momega^(2)A^(2)` `A^(2)-x^(2)=(3)/(4)A^(2)` `A^(2)-(3)/(4)A^(2)=x^(2)` `(A^(2))/(4)=x^(2)` `therefore x=(A)/(2)` |
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