The motion of a particle represented by y = s | vedclass

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(B) Given equation: $y = \sin \omega t - \cos \omega t$.Multiply and divide by $\sqrt{2}$:$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \fra

Vedclass · Accepted Answer

$S.H.M.$ with period $\frac{2\pi}{\omega}$,amplitude $\sqrt{2}$

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(B) Given equation: $y = \sin \omega t - \cos \omega t$.Multiply and divide by $\sqrt{2}$:$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} \sin \omega t - \frac{1}{\sqrt{2}} \cos \omega t \right)$.Using the trigonometric identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we can write:$y = \sqrt{2} \sin \left( \omega t - \frac{\pi}{4} \right)$.This is the standard form of a Simple Harmonic Motion ($S$.$H$.$M$.) equation,$y = A \sin(\omega t + \phi)$,where $A$ is the amplitude and $\omega$ is the angular frequency.Comparing the equations,the amplitude $A = \sqrt{2}$.The time period $T$ is given by $T = \frac{2\pi}{\omega}$.Thus,the motion is $S.H.M.$ with period $\frac{2\pi}{\omega}$ and amplitude $\sqrt{2}$.

The motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is

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