MHTCET 14 May 2023 Shift 1 Solved Paper

The given wave equation is:$Y=10\sin\left(\frac{2\pi t}{30}+\alpha\right)$At $t=0$, the displacement $Y$ is given as $5\text{ cm}$. So, let's find the phase $\alpha$ using this information:$5=10\sin(\alpha)$$\frac{5}{10}=\frac{1}{2}=\sin(\alpha)$ Since $\sin(30^{\circ})=\frac{1}{2}$, we can say that $\alpha=30^{\circ}$ or $\alpha=\frac{\pi}{6}$ radians, considering that sine is positive in the first and second quadrants, and seeing that $\alpha$ represents a phase shift, it usually takes the smallest positive angle that satisfies the equation.Now, let's calculate the total phase at $t=7.5\text{ s}$ :Total phase $=\frac{2\pi t}{30}+\alpha$Plug in the values:Total phase $=\frac{2\pi\cdot7.5}{30}+\frac{\pi}{6}$Total phase $=\frac{2\pi\cdot1}{4}+\frac{\pi}{6}=\frac{\pi}{2}+\frac{\pi}{6}$ To add these fractions, we first find a common denominator :Total phase $=\frac{3\pi}{6}+\frac{\pi}{6}=\frac{4\pi}{6}$Now we simplify the fraction:Total phase $=\frac{2\pi}{3}$Therefore, the total phase at $t=7.5\text{ s}$ is $\frac{2\pi}{3}$ radians, which corresponds to Option D.