$A$ 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 $s$)?

$A$ 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

$A$ string has a mass per unit length of $10^{-6} \,kg/cm$. The equation of a simple harmonic wave produced in it is $Y=0.2 \sin(2x+80t) \,m$. The tension in the string is: (in $N$)

Two strings with circular cross section and made of same material are stretched to have the same amount of tension. $A$ transverse wave is then made to pass through both the strings. The velocity of the wave in the first string having the radius of cross section $R$ is $v_1$,and that in the other string having radius of cross section $R/2$ is $v_2$. Then $\frac{v_2}{v_1} = $

$A$ uniform rope of mass $0.1 \,kg$ and length $2.45 \,m$ hangs from a rigid support. The time taken by a transverse wave formed in the rope to travel through the full length of the rope is (Assume $g = 9.8 \,m/s^2$). (in $\,s$)

$A$ uniform metal wire of length $L$,mass $M$,and density $\rho$ is under a tension $T$. If the speed of a transverse wave along the wire is $V$,then the area of cross-section of the wire is: