The equation of a wave is given by y = 10 sin | vedclass

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(B) The given wave equation is $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha ight)$.At $t = 0$,the displacement $y = 5 	ext{ cm}$.Substituting th

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$\frac{\pi}{2}$

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(B) The given wave equation is $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha ight)$.At $t = 0$,the displacement $y = 5 	ext{ cm}$.Substituting these values: $5 = 10 \sin \left( \frac{2 \pi}{45} 	imes 0 + \alpha ight) = 10 \sin \alpha$.Thus,$\sin \alpha = \frac{5}{10} = \frac{1}{2}$,which gives $\alpha = \frac{\pi}{6}$.The total phase of the wave is given by $\phi = \frac{2 \pi}{45} t + \alpha$.At $t = 7.5 	ext{ s} = \frac{15}{2} 	ext{ s}$,the total phase is:$\phi = \frac{2 \pi}{45} 	imes \frac{15}{2} + \frac{\pi}{6} = \frac{\pi}{3} + \frac{\pi}{6}$.$\phi = \frac{2 \pi + \pi}{6} = \frac{3 \pi}{6} = \frac{\pi}{2}$.

The equation of a wave is given by $y = 10 \sin \left( \frac{2 \pi}{45} t + \alpha \right)$. If the displacement is $5 \text{ cm}$ at $t = 0$,then the total phase at $t = 7.5 \text{ s}$ is

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