A long composite string is made up by joining two strings of different mass per unit length  $\mu \text{\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}}4\mu$ respectively. The composite string is under the same tension. A transverse wave pulse $Y=\left(6\text{\hspace{0.17em}mm}\right)\sin \left(5t+40x\right),$ 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

Given wave pulse  $Y=\left(6\text{\hspace{0.17em}mm}\right)\sin \left(5t+40x\right)\mathrm{......}\left(1\right)$
We know speed of wave $v=\sqrt{\frac{T}{\mu }}\mathrm{....}\left(2\right)$
Assume  $v_1=$Speed in string 1 (of density ${\mu }_1$ )
$v_2=$Speed in string 2 (of density  ${\mu }_2$)

$v_1=\sqrt{\frac{T}{\mu }};v_2=\sqrt{\frac{T}{4\mu }}$

$\Rightarrow \frac{v_1}{v_2}=2$

$\therefore v_2<v_1$

$\Rightarrow$ 2nd medium is denser
 $\therefore$ The wave reflected from the denser medium has phase change of   Amplitude of reflected wave is
$\Rightarrow A_r=\frac{v_2-v_1}{v_2+v_1}\times A_i$ 
 Amplitude of incident wave
Substituting the values we get
$\Rightarrow A_r=\frac{v_2-v_1}{v_2+v_1}\times 6=\frac{\frac{v_1}{2}-v_1}{\frac{v_2}{2}+v_1}\times 6=-2\text{\hspace{0.17em}mm}$ 
$\therefore \text{equation\hspace{0.17em}of\hspace{0.17em}reflected wave pulse is\hspace{0.17em}}Y=\left(-2\text{\hspace{0.17em}mm}\right)\sin \left(5t-40x\right)$
Therefore, the correct answer is (C).