question_answer

A composite string is made up by joining two strings of different masses per unit length u, and 4u.The composite string is under the same tension. A transverse wave pulse 7= (6 mm) sin (5t + 40 x), where 't' is in seconds and 'x' is in metres, is sent along the lighter string towards the joint. The joint is at x = 0. The equation of the wave pulse reflected from the joint is

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 \[Y=-(2mm)\sin (5t-40x)\]