The displacement of a particle performing S.H | vedclass

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(C) The displacement equation is $y = A \cos(\pi t + \pi \phi)$.Taking the derivative with respect to time,the velocity is $v = \frac{dy}{dt} = -A\pi

Vedclass · Accepted Answer

$2\sqrt{2}$

Vedclass · Answer

(C) The displacement equation is $y = A \cos(\pi t + \pi \phi)$.Taking the derivative with respect to time,the velocity is $v = \frac{dy}{dt} = -A\pi \sin(\pi t + \pi \phi)$.At $t = 0$,we have $y_0 = A \cos(\pi \phi)$ and $v_0 = -A\pi \sin(\pi \phi)$.From these,we get $\cos(\pi \phi) = \frac{y_0}{A}$ and $\sin(\pi \phi) = -\frac{v_0}{A\pi}$.Using the identity $\cos^2(\pi \phi) + \sin^2(\pi \phi) = 1$,we get $\left(\frac{y_0}{A}ight)^2 + \left(-\frac{v_0}{A\pi}ight)^2 = 1$.This simplifies to $y_0^2 + \frac{v_0^2}{\pi^2} = A^2$.Substituting the given values $y_0 = 2 	ext{ cm}$ and $v_0 = 2\pi 	ext{ cm/s}$:$2^2 + \frac{(2\pi)^2}{\pi^2} = A^2$.$4 + 4 = A^2$.$A^2 = 8$.$A = \sqrt{8} = 2\sqrt{2} 	ext{ cm}$.

The displacement of a particle performing $S.H.M.$ is given by $Y = A \cos [\pi(t + \phi)]$. If at $t = 0$,the displacement is $y = 2 \text{ cm}$ and velocity is $v = 2\pi \text{ cm/s}$,the value of amplitude $A$ in $\text{cm}$ is:

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