Question
The phase difference between two $SHM\,\,$ ${y_1}\, = \,10\,\sin \,\left( {10\pi t\, + \,\frac{\pi }{3}} \right)$ and ${y_2}\, = \,12\,\sin \,\left( {8\pi t\, + \,\frac{\pi }{4}} \right)$ at $t = 0.5\,s$ it
✓
Answer
Phase for $1\, st$ $\mathrm{SHM}: \phi_{1}=10 \pi \mathrm{t}+\frac{\pi}{3}$
Phase for $2 \,nd$ $\mathrm{SHM}: \phi_{2}=8 \pi \mathrm{t}+\frac{\pi}{4}$
Phase difference : $\Delta \phi=\phi_{1}-\phi_{2}=2 \pi t+\frac{\pi}{12}$
Put $t=0.5 \mathrm{s}$
$\Delta \phi=2 \pi(0.5)+\frac{\pi}{12}=\frac{13 \pi}{12}$
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