A) \[y=a\sin (\omega t+kx)\]
B) \[y=-a\sin (\omega t+kx)\]
C) \[y=a\sin (\omega t-kx)\]
D) \[y=-a\sin (\omega t-kx)\]
Correct Answer: B
Solution :
Equation of a wave \[y=a\sin (\omega t-kx)\] ...(1) Let equations of another wave may be, \[y=a\sin (\omega t+kx)\] ...(2) \[y=-a\sin (\omega t+kx)\] ...(3) If eq. (1) propagates with eq. (2), then we get \[y=2a\cos kx\sin \omega t\] ...(4) If eq. (1), propagates with eq. (3), then we get \[y=-2a\sin kx\cos \omega t\] ...(5) After putting\[x=0\]in eqs (4) and (5) respectively, we get \[y=2a\sin \omega t\] and \[y=0\] Hence, eq. (3) is a equation of unknown wave.You need to login to perform this action.
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