$A$ perfectly elastic uniform string is suspended vertically with its upper end fixed to the ceiling and the lower end loaded with a weight. If a transverse wave is imparted to the lower end of the string,the pulse will

$A$ mass of $20\ kg$ is hanging with the support of two strings of the same linear mass density. Now,pulses are generated in both strings at the same time near the joint at the mass. The ratio of the time taken by a pulse to travel through string $1$ to that taken by a pulse on string $2$ is:

$A$ 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:

$A$ transverse wave travels on a taut steel wire with a velocity of $v$ when the tension in it is $2.06 \times 10^{4} \; N$. When the tension is changed to $T$,the velocity changes to $v/2$. The value of $T$ is close to:

In the figure shown,a mass of $1 \ kg$ is connected to a string of mass per unit length $1.2 \ g/m$. The length of the string is $1 \ m$ and its other end is connected to the ceiling of a lift which is accelerating upwards with an acceleration of $2 \ m/s^2$. $A$ transverse pulse is produced at the lowest point of the string. The time taken by the pulse to reach the top of the string is .... $s$. (Take $g = 10 \ m/s^2$)