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(C) The velocity of a particle in simple harmonic motion at the mean position is given by $V_{max} = A\omega$.Given $A = 4 \ cm$ and $V_{max} = 10 \ c

Vedclass · Accepted Answer

$12$

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(C) The velocity of a particle in simple harmonic motion at the mean position is given by $V_{max} = A\omega$.Given $A = 4 \ cm$ and $V_{max} = 10 \ cm/s$,we have $10 = 4\omega$,which gives $\omega = 2.5 \ rad/s$.The velocity $V$ at any displacement $x$ from the mean position is given by $V = \omega \sqrt{A^2 - x^2}$.Substituting the given values: $5 = 2.5 \sqrt{4^2 - x^2}$.Dividing by $2.5$: $2 = \sqrt{16 - x^2}$.Squaring both sides: $4 = 16 - x^2$.Therefore,$x^2 = 12$,which means $x = \sqrt{12} \ cm$.Comparing this with $\sqrt{\alpha} \ cm$,we get $\alpha = 12$.

$A$ particle executes simple harmonic motion with an amplitude of $4 \ cm$. At the mean position,the velocity of the particle is $10 \ cm/s$. The distance of the particle from the mean position when its speed becomes $5 \ cm/s$ is $\sqrt{\alpha} \ cm$,where $\alpha = $ . . . . . . .

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