The initial phase of a body executing SHM is | vedclass

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(A) The phase of a particle executing $SHM$ at any time $t$ is given by $	heta = \omega t + \phi$,where $\phi$ is the initial phase.One complete osci

Vedclass · Accepted Answer

$\frac{81\pi}{4} 	ext{ rad}$

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(A) The phase of a particle executing $SHM$ at any time $t$ is given by $	heta = \omega t + \phi$,where $\phi$ is the initial phase.One complete oscillation corresponds to a phase change of $2\pi$ radians.After $n$ oscillations,the total phase change is $2\pi n$.Given that the number of oscillations $n = 10$ and the initial phase $\phi = \frac{\pi}{4}$.The final phase $	heta$ is given by $	heta = 2\pi n + \phi$.Substituting the values: $	heta = 2\pi(10) + \frac{\pi}{4}$.$	heta = 20\pi + \frac{\pi}{4} = \frac{80\pi + \pi}{4} = \frac{81\pi}{4} 	ext{ rad}$.

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

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