The displacement of a particle executing SHM | vedclass

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(B) The displacement equation for a particle executing $SHM$ is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 3 \sin \left(

Vedclass · Accepted Answer

$3 \ m, 6 \pi \ m/s$

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(B) The displacement equation for a particle executing $SHM$ is given by $x = A \sin(\omega t + \phi)$.Comparing the given equation $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}ight)$ with the standard form:Amplitude $A = 3 \ m$.Angular frequency $\omega = 2 \pi \ rad/s$.The maximum speed $v_{\max}$ of a particle in $SHM$ is given by the formula $v_{\max} = \omega A$.Substituting the values: $v_{\max} = (2 \pi \ rad/s) 	imes (3 \ m) = 6 \pi \ m/s$.Thus,the amplitude is $3 \ m$ and the maximum speed is $6 \pi \ m/s$.

The displacement of a particle executing $SHM$ is given by $x = 3 \sin \left(2 \pi t + \frac{\pi}{4}\right)$,where $x$ is in $m$ and $t$ is in $s$. The amplitude and maximum speed of the particle are:

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