Q137 / 180 · KCET 2011 · Physics

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

  • A
    $\frac{\pi}{3}$
  • Correct answer
    B
    $\frac{\pi}{2}$
  • C
    $\frac{\pi}{6}$
  • D
    $\pi$

Solution

$y=10 \sin \left[\frac{2 \pi}{45} t+\alpha\right]$ $y=5 \mathrm{~cm}$ at $t=0$ $\therefore \quad 5=10(\sin \alpha)$ or $\sin \alpha=\frac{1}{2}$ or $\alpha=\frac{\pi}{6}$ Again if $t=7.5 \mathrm{~s}$ Then total phase $=\frac{2 \pi}{45} \times \frac{15}{2}+\frac{\pi}{6}$ $=\frac{\pi}{3}+\frac{\pi}{6}$ $=\frac{2 \pi+\pi}{6}=\frac{3 \pi}{6}=\frac{\pi}{2}$

Hence, the correct answer is option: 'B'.

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