MCQ
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
  • A
    $\frac {11 \pi}{12}$
  • $\frac {13 \pi}{12}$
  • C
    $\pi $
  • D
    $\frac {17 \pi}{12}$

Answer

Correct option: B.
$\frac {13 \pi}{12}$
b
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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