Wave Mechanics Question 663
Question: 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
Options:
A) $ Y=(2mm)\sin (5t-40x) $
B) $ Y=(4mm)\sin (40x-5t) $
C) $ Y=-(2mm)\sin (5t-40x) $
D) $ Y=(2mm)\sin (5t-10x) $
Show Answer
Answer:
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 ,eq^{n} $ of reflected wave pulse is $ Y=-(2mm)\sin (5t-40x) $
BETA
