The displacement of a particle in S.H.M. vari | vedclass

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(C) The given equation is $x = 4(\cos \pi t + \sin \pi t)$.We can rewrite this in the form $x = A \sin(\omega t + \phi)$ or $x = A \cos(\omega t + \ph

Vedclass · Accepted Answer

$4 \sqrt{2}$

Vedclass · Answer

(C) The given equation is $x = 4(\cos \pi t + \sin \pi t)$.We can rewrite this in the form $x = A \sin(\omega t + \phi)$ or $x = A \cos(\omega t + \phi)$.Multiply and divide by $\sqrt{4^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}$.$x = 4\sqrt{2} \left( \frac{4}{4\sqrt{2}} \cos \pi t + \frac{4}{4\sqrt{2}} \sin \pi t \right)$.$x = 4\sqrt{2} \left( \frac{1}{\sqrt{2}} \cos \pi t + \frac{1}{\sqrt{2}} \sin \pi t \right)$.Using the trigonometric identity $\sin(A + B) = \sin A \cos B + \cos A \sin B$,we can write:$x = 4\sqrt{2} \sin(\pi t + \frac{\pi}{4})$.Comparing this with the standard equation $x = A \sin(\omega t + \phi)$,the amplitude $A$ is $4\sqrt{2}$.

The displacement of a particle in $S.H.M.$ varies according to the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is

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