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))

A uniform rope with length L and mass m is held at one end and whirled

1.	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» SOLUTION :`-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)`