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(C) The given displacement equation is $x = 4(\cos \pi t + \sin \pi t)$.To find the amplitude, we express the equation in the form $x = A \sin(\omega

Vedclass · Accepted Answer

$4 \sqrt{2}$

Vedclass · Answer

(C) The given displacement equation is $x = 4(\cos \pi t + \sin \pi t)$.To find the amplitude, we express the equation in the form $x = A \sin(\omega t + \phi)$.Multiply and divide by $\sqrt{1^2 + 1^2} = \sqrt{2}$:$x = 4 \sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$, where $\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$ and $\cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$:$x = 4 \sqrt{2} \left( \sin \frac{\pi}{4} \cos \pi t + \cos \frac{\pi}{4} \sin \pi t \right)$.$x = 4 \sqrt{2} \sin \left( \pi t + \frac{\pi}{4} \right)$.Comparing this with the standard $SHM$ equation $x = A \sin(\omega t + \phi)$, the amplitude $A$ is $4 \sqrt{2}$.

The displacement of a particle is given by the relation $x = 4(\cos \pi t + \sin \pi t)$. The amplitude of the particle is

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