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(A) The given wave equation is $y = 5 	imes 10^{-3} \sin \left(12.5 \pi x - \frac{\pi}{2} tight)$.Comparing this with the standard wave equation $y

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$0.16 \ m, 4 \ s$

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(A) The given wave equation is $y = 5 	imes 10^{-3} \sin \left(12.5 \pi x - \frac{\pi}{2} tight)$.Comparing this with the standard wave equation $y = A \sin(kx - \omega t)$, we get:Wave number $k = 12.5 \pi$ and angular frequency $\omega = \frac{\pi}{2}$.We know that $k = \frac{2 \pi}{\lambda}$, so $\lambda = \frac{2 \pi}{k} = \frac{2 \pi}{12.5 \pi} = \frac{2}{12.5} = 0.16 \ m$.We also know that $\omega = \frac{2 \pi}{T}$, so $T = \frac{2 \pi}{\omega} = \frac{2 \pi}{\pi / 2} = 4 \ s$.Therefore, the wavelength is $0.16 \ m$ and the time period is $4 \ s$.

$A$ wave is given by $y=5 \times 10^{-3} \sin \left(12.5 \pi x - \frac{\pi}{2} t\right)$. Then its wavelength and time period are respectively ($y$ and $x$ are in metres and $t$ is in seconds).

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