# A light string of mass m and length L has its ends tied to two walls that are separated by the distance D . Two objects each of mass M are suspended from the string as shown in figure given . if a wave pulse is sent from point A find speed is AB string

$(a)\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\qquad(b)\;\sqrt{\large\frac{(2D-L)}{4D^2-4DL}}\qquad(c)\;\sqrt{2l(2D-L)}\qquad(d)\;None$

Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$
Explanation :
$2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$
$1+cos \theta =\large\frac{2D}{L}$
$cos \theta=\large\frac{2D-L}{L}$
$Tsin \theta =mg$
$T=\large\frac{mg}{sin \theta}$
$T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$
$v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$
$v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$
$v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$
$v=\sqrt{cos \theta 2l}$
$=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$
Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$
Explanation :
$2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$
$1+cos \theta =\large\frac{2D}{L}$
$cos \theta=\large\frac{2D-L}{L}$
$Tsin \theta =mg$
$T=\large\frac{mg}{sin \theta}$
$T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$
$v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$
$v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$
$v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$
$v=\sqrt{cos \theta 2l}$
$=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$
Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$
Explanation :
$2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$
$1+cos \theta =\large\frac{2D}{L}$
$cos \theta=\large\frac{2D-L}{L}$
$Tsin \theta =mg$
$T=\large\frac{mg}{sin \theta}$
$T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$
$v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$
$v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$
$v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$
$v=\sqrt{cos \theta 2l}$
$=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$
Answer : (a) $\;\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}$
Explanation :
$2 \times \large\frac{L}{4} cos \theta + \large\frac{L}{2}=D$
$1+cos \theta =\large\frac{2D}{L}$
$cos \theta=\large\frac{2D-L}{L}$
$Tsin \theta =mg$
$T=\large\frac{mg}{sin \theta}$
$T cos \theta =T_{1} \qquad \;$ AB string mass = $\large\frac{m}{2}$
$v=\sqrt{\large\frac{T_{1}}{\mu}}= \qquad \; \mu=\large\frac{mg}{2l}$
$v=\sqrt{\large\frac{T cos \theta 2l}{mg}}$
$v=\sqrt{\large\frac{mg}{sin \theta} \times \large\frac{cos 2 \theta 2l}{mg}}$
$v=\sqrt{cos \theta 2l}$
$=\sqrt{\large\frac{2l(2D-L)}{\sqrt{4D^2-4DL}}}\;.$