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(D) The given equation is $y = 5 \sin \pi (t + 4) = 5 \sin (\pi t + 4\pi)$.Comparing this with the standard simple harmonic motion equation $y = a \si

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$a = 5, T = 2$

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(D) The given equation is $y = 5 \sin \pi (t + 4) = 5 \sin (\pi t + 4\pi)$.Comparing this with the standard simple harmonic motion equation $y = a \sin (\omega t + \phi)$ or $y = a \sin (\frac{2\pi}{T} t + \phi)$:The amplitude $a$ is the coefficient of the sine function,so $a = 5 \ m$.The angular frequency $\omega$ is the coefficient of $t$,so $\omega = \pi \ rad/s$.Since $\omega = \frac{2\pi}{T}$,we have $\pi = \frac{2\pi}{T}$.Solving for $T$,we get $T = 2 \ s$.

$A$ simple harmonic wave having an amplitude $a$ and time period $T$ is represented by the equation $y = 5 \sin \pi (t + 4) \ m$. Then the value of amplitude $(a)$ in $(m)$ and time period $(T)$ in seconds are:

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