Question

a uniform rope of mass 0.1kg and length 2.45m hangs from the ceiling find speed of the transverse wave in the rope at a point 0.5 m distance from the lower end

Answer 1

solution:

mass=m=0.2kg
g=9.8m/s²
length=2.45m
h=0.5mT
ension at a point h meter height from the lower end of the rope=mgh

speed of transverse wave V=√(T/m)
=√[mgh/m]
=√gh
Substituting values:
V=√[9.8x0.5)=2.21 m/s

b]At lower end of rope, tension is zero hence speed of transverse wave is zerospeed at upper end =√gl
acceleration=√gl / t

l=1/2√[gl]xt²/t
t=2√[l/g]
=2x√[2.45/9.8]=1sec

Answer 2

Let a uniform rope of length L and mass m is given as shown in figure .
we cut an elementary length of dx form the lower end of rope.
so, mass of element = $\bold{\frac{m}{L}dx}$
Tension on elementary part = mg.dx/L

Now, formula of velocity is $\bold{\sqrt{\frac{T}{\mu}}}$
Where T is tension in string e.g., T = mg.dx/L
μ is mass per unit length e.g., mass per unit length ( μ ) =( m.dx/L)/x = m.dx/Lx
∴ velocity of wave at x from the lower point = $\bold{\sqrt{\frac{mg.dx/L}{m.dx/Lx}}}$ $= \bold{\sqrt{gx}}$

we get the formula, for finding velocity of wave at x from the lower end of the rope is √(gx)

Here, g = 9.8 m/s² and x = 0.5 m
v = √{9.8 × 0.5} = √{4.9} = 2.21 m/s
Hence, andwer is 2.21 m/s