$A$ string of length $L$ and mass $M$ hangs freely from a fixed point. The velocity of transverse waves along the string at a distance $x$ from the free end is

When the area of cross-section of a stretched wire is halved and tension is doubled,the speed of propagation of transverse waves along it becomes $k$ times the initial speed. Then,$k$ is:

Two strings $(A, B)$ having linear densities $\mu_{A} = 2 \times 10^{-4} \ kg/m$ and $\mu_{B} = 4 \times 10^{-4} \ kg/m$ and lengths $L_{A} = 2.5 \ m$ and $L_{B} = 1.5 \ m$ respectively are joined. Free ends of $A$ and $B$ are tied to two rigid supports $C$ and $D$, respectively, creating a tension of $500 \ N$ in the wire. Two identical pulses, sent from $C$ and $D$ ends, take time $t_1$ and $t_2$, respectively, to reach the joint. The ratio $t_1 / t_2$ is:

Obtain the equation of speed of transverse wave on a tensed (stretched) string.

$A$ uniform wire $20 \,m$ long and weighing $50 \,N$ hangs vertically. The speed of the wave at the midpoint of the wire is (acceleration due to gravity $= g = 10 \,ms^{-2}$)

$A$ composite string is made up by joining two strings of different mass per unit length,$\mu$ and $4\mu$. The composite string is under the same tension $T$. $A$ transverse wave pulse,$Y = (6 \text{ mm}) \sin(5t + 40x)$,where $t$ is in seconds and $x$ is in meters,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: