A) 1.25 rad
B) 1.57 rad
C) 0.57 rad
D) 1.0 rad
Correct Answer: D
Solution :
| \[{{y}_{1}}=a\sin (\omega t+kx+0.57)m\] |
| and \[{{y}_{2}}=a\cos (\omega t+kx)m\] |
| or \[{{y}_{2}}=a\sin \left( \frac{\pi }{2}+\omega t+kx \right)m\] |
| Phase difference |
| \[\Delta \phi ={{\phi }_{2}}-{{\phi }_{1}}\] |
| \[=\frac{\pi }{2}-0.57\] |
| \[=1.57-0.57\] |
| \[=1\,\,\text{rad}\] |
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