Question

a pulse is generated at lower end of a hanging rope of uniform density and length L the speed of the process when it reaches the midpoint of rope is

Answer 1

Let a pulse is generated at lower end of a hanging rope of uniforms density and length L .

speed of wave , $v=\sqrt{\frac{T}{\mu}}$

where T is tension and $\mu$ is linear mass density.

Cut an element of thickness, dx , x distance from lower end of hanging rope. so, mass of rope of length x , $dm=\left(\frac{mx}{l}\right)$

$T=\left(\frac{mgx}{l}\right)$

now, velocity of wave at x distance from lower end , $v=\sqrt{\frac{mgx/l}{m/l}}=\sqrt{gx}$

The speed of the process when it reaches the midpoint of the rope is v = $\sqrt{gL/2}$

Answer 2

Answer:

Let a pulse is generated at lower end of a hanging rope of uniforms density and length L .

speed of wave , v=\sqrt{\frac{T}{\mu}}v=

μ

T

where T is tension and \muμ is linear mass density.

Cut an element of thickness, dx , x distance from lower end of hanging rope. so, mass of rope of length x , dm=\left(\frac{mx}{l}\right)dm=(

l

mx

)

T=\left(\frac{mgx}{l}\right)T=(

l

mgx

)

now, velocity of wave at x distance from lower end , v=\sqrt{\frac{mgx/l}{m/l}}=\sqrt{gx}v=

m/l

mgx/l

=

gx

The speed of the process when it reaches the midpoint of the rope is v = \sqrt{gL/2}

gL/2