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:

$A$ transverse wave propagating on a stretched string of linear density $3 \times 10^{-4} \ kg \ m^{-1}$ is represented by the equation $y = 0.2 \sin (1.5 x + 60 t)$, where $x$ is in metres and $t$ is in seconds. The tension in the string (in newton) is

$A$ wire of length $50 \ cm$ and weighing $10 \ g$ is attached to a spring at one end and to a fixed wall at the other end. The spring has a spring constant of $50 \ N/m$ and is stretched by $1 \ cm$. If a wave pulse is produced on the string near the wall, then how much time will it take to reach the spring (in $s$)?

$A$ uniform string is suspended vertically. $A$ transverse pulse is created at the top of the string. Then:

When the length of a tense wire is made half while keeping its mass constant,what will be the effect on the speed of the transverse wave in it?

$A$ string of mass $0.1 \, kg$ is under a tension $1.6 \, N$. The length of the string is $1 \, m$. $A$ transverse wave starts from one end of the string. The time taken by the wave to reach the other end is (in $s$.)