A uniform rope of length L and mass m, hangs vertically from a rigid support. A block of mass `m_(2)` is attached to the free end of the rope. A transverse pulse of wavelength`lamda_(1)` is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is `lamda_(2).` . The ratio `lamda_(2)//lamda_(1)` is ......

Options

A. `sqrt (( m _(1) + m_(2))/( m _(2)))`
B. `sqrt ((m _(2))/( m _(1)))`
C. `sqrt (( m _(1) + m _(2))/( m _(1)))`
D. `sqrt (( m _(1))/( m _(2)))`

Answer

A

Explanation

(i) At the lower end of wire, `T_(1) = m _(2) g `
(ii) At the upper end of wire, `T_(2) = (m _(1) + m _(2)) g `
Now, we have `v = f lamda = sqrt ((T)/(mu))`
`therefore v _(1) = f lamda_(1) = sqrt ((T_(1))/( mu))`
`v _(2) = f lamda_(2) = sqrt ((T_(2))/( mu ))`
Taking ratio, `(lamda _(1))/( lamda _(2)) = sqrt (( T _(1))/( T _(2)) ) = sqrt (( m _(2) g )/( (m _(1) + m _(2)) g ))`
`therefore (lamda _(2))/( lamda _(1)) = sqrt (( m _(1) + m _(2))/( m _(2)))`