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 \[y=a\sin (\omega \,t-kx)\] ... (i) Let equations of another wave may be \[y=a\sin (\omega \,t+kx)\] ... (ii) \[y=-a\sin (\omega \,t+kx)\] ... (iii) If Eq. (i) propagates with Eq. (ii), then we get the resultant wave \[y=2\,a\cos \,kx\sin \omega \,t\] ... (iv) If Eq. (i) propagates with Eq. (iii), then we get \[y=-2a\,\sin kx\,\cos \omega t\] ?. (v) After putting \[x=0\] in Eqs. (iv) and (v) respectively, we get \[y=2\,a\,\sin \omega \,t\] and \[y=0\] Hence, Eq. (iii) is a equation of unknown wave.
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