A uniform rope with length `L` and mass `m` is held at one end and whirled in a horizontal circle with angular velocity `omega`. You can ignor the force of gravity on the rope. Find the time required for a transverse wave to travel from one end of the rope to the other.
Hint: `int (dx)/(sqrt(a^(2) - (x^(2)))) = sin^(-1)((x)/(a))`

Answer

A, B

Explanation

`-dT = (dm)xomega^(2) =((m)/(L)dx)xomega^(2)`
or `-int_(0)^(T)dT = (m)/(L)omega^(2) int_(x = L)^(x)x dx`
`:. -T = (m)/(L)omega^(2) ((x^(2))/(2) - (L^(2))/(2))`
or `T = (momega^(2))/(2L)(L^(2) - x^(2))`
`v=sqrt((T)/(mu))`
sqrt(((momega^(2))/(2L)(L^(2) - x^(2)))/(m//L))`
`=omegasqrt((L^(2) - x^(2))/(2))`
or `(dx)/(dt) = (omega)/(sqrt(2))sqrt(L^(2) - x^(2))`
`:. int_(0)^(t) dt = sqrt(2)/(omega)int_(0)^(L)(dx)/(sqrt(L^(2) - x^(2)))`
`:. t = sqrt(2)/(omega)[sin^(-1)((x)/(L))]_(0)^(L)`
`=sqrt(2)/(omega)*(pi)/(2) = (pi)/(sqrt2omega)`