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 wavelengthlamda_(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 ......

A uniform rope of length L and mass m, hangs vertically from a rigid s

1.	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 wavelengthlamda_(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 ......
Answer» `sqrt (( m _(1) + m_(2))/( m _(2)))` `sqrt ((m _(2))/( m _(1)))` `sqrt (( m _(1) + m _(2))/( m _(1)))` `sqrt (( m _(1))/( m _(2)))` Solution :(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)))`