A) \[Y=(2mm)\sin (5t-40x)\]
B) \[Y=(4mm)\sin (40x-5t)\]
C) \[Y=-(2mm)\sin (5t-40x)\]
D) \[Y=(2mm)\sin (5t-10x)\]
Correct Answer: C
Solution :
[c] \[{{v}_{1}}=\sqrt{\frac{T}{\mu }};\,\,\,{{v}_{2}}=\sqrt{\frac{T}{4\mu }}\] \[{{v}_{2}}<{{v}_{1}}\Rightarrow 2nd\]medium is denser \[\therefore \] The wave reflected from the denser medium has phase change of\[\pi \]. \[\Rightarrow \,{{A}_{r}}=\frac{{{v}_{2}}-{{v}_{1}}}{{{v}_{2}}+{{v}_{1}}}\times 6=\frac{\frac{{{v}_{1}}}{2}-{{v}_{1}}}{\frac{{{v}_{2}}}{2}+{{v}_{1}}}\times 6=-2mm\] \[\Rightarrow \,e{{q}^{n}}\]of reflected wave pulse is \[Y=-(2mm)\sin (5t-40x)\]You need to login to perform this action.
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