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(D) The time period of a simple harmonic motion $(SHM)$ is given by $T = \frac{2\pi}{\omega'}$,where $\omega'$ is the angular frequency of the $SHM$.G

Vedclass · Accepted Answer

$3 \cos \left(\frac{\pi}{4}-2 \omega t\right)$

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(D) The time period of a simple harmonic motion $(SHM)$ is given by $T = \frac{2\pi}{\omega'}$,where $\omega'$ is the angular frequency of the $SHM$.Given $T = \frac{\pi}{\omega}$,we have $\frac{\pi}{\omega} = \frac{2\pi}{\omega'}$,which implies $\omega' = 2\omega$.We need to find the function that represents $SHM$ with an angular frequency of $2\omega$.For option $(C)$: $\sin^2(\omega t) = \frac{1 - \cos(2\omega t)}{2} = \frac{1}{2} - \frac{1}{2}\cos(2\omega t)$. This represents a constant shift plus an $SHM$ with angular frequency $2\omega$.For option $(D)$: $3 \cos \left(\frac{\pi}{4} - 2\omega t\right) = 3 \cos \left(2\omega t - \frac{\pi}{4}\right)$. This is a standard $SHM$ equation $x(t) = A \cos(\omega' t + \phi)$ with angular frequency $\omega' = 2\omega$.Both $(C)$ and $(D)$ have the correct frequency,but $(D)$ is a pure $SHM$ function,whereas $(C)$ includes a constant term.

The function of time representing a simple harmonic motion with a period of $\frac{\pi}{\omega}$ is :

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