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(C) The displacement function is $x(t) = A \sin (\omega t + \phi)$.At $t = 0$,$x(0) = A \sin \phi = 2 \dots (1)$.The velocity function is $v(t) = \fra

Vedclass · Accepted Answer

$2$

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(C) The displacement function is $x(t) = A \sin (\omega t + \phi)$.At $t = 0$,$x(0) = A \sin \phi = 2 \dots (1)$.The velocity function is $v(t) = \frac{dx}{dt} = A \omega \cos (\omega t + \phi)$.At $t = 0$,$v(0) = A \omega \cos \phi = 2 \omega \implies A \cos \phi = 2 \dots (2)$.Dividing equation $(1)$ by equation $(2)$:$\frac{A \sin \phi}{A \cos \phi} = \frac{2}{2} \implies 	an \phi = 1 \implies \phi = 45^{\circ}$.Substituting $\phi = 45^{\circ}$ in equation $(1)$:$A \sin 45^{\circ} = 2 \implies A \left( \frac{1}{\sqrt{2}} ight) = 2 \implies A = 2 \sqrt{2} \, cm$.Comparing $A = 2 \sqrt{2} \, cm$ with $x \sqrt{2} \, cm$,we get $x = 2$.

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