A stone is tied to an elastic string of negligible mass and spring constant k. The unstretched length of the string is L and has negligible mass. The other end of the string is fixed to a nail at a point P. Initially the stone is at the same level as the point P. The stone is dropped vertically from point P.
(a) Find the distance 'y' from the top when the mass comes to rest for an instant, for the first time.
(b) What is the maximum velocity attained by the stone in this drop ?
(c) What shall be the nature of the motion after the stone has reached its lowest point ?

Explanation

Consider the diagram, the stone is dropped from point P.

Stone is in free fall upto length L. After that elasticity of string exerted force for SHM. Suppose, stone is at rest at instantaneous distance .y..
Loss in potential energy of stone = Gain in elastic potential energy in string.
`mgy = 1/2 k (y-L) ^(2)`
`therefore mgy = 1/2 ky ^(2) -kyL + 1/2 kL^(2)`
`implies1/2 ky ^(2) - (kL +mg) + 1/2 kL ^(2) =0`
`y = ((kL + mg) pm sqrt ((kL +mg) ^(2) -k^(2) L ^(2)))/( k )`
`therefore y = ((k L +mg ) pm sqrt ( 2 mg k L + m ^(2) g ^(2)))/( k )`
`therefore y = ((k L + mg ) + sqrt (2m k L + m^(2) g ^(2)))/( k )`
(Taking .+. sign only)
(b) In SHM, the maximum velocity is attained when the body passes, through the "equilibrium positon". Means, at that time the instantaneous acceleration is zero. That is `mg - kx =0,` where x is the extension from L, `therefore mg = kx`
If v is the velocity, then
`1/2 mv ^(2) + 1/2 kx ^(2) = mg ( L +x)`
(From the law of conservation of energy)
`1/2 mv ^(2) = mg (L+x) -1/2 kx ^(2)`
Now `mg = kx implies x = (mg )/(k)`
`therefore 1/2 mv ^(2) =mg (L + (mg )/(k)) -1/2 k (m ^(2) g ^(2))/( k)= mgL + (m^(2) g ^(2))/(k ) - 1/2 ( m ^(2) g ^(2))/(k)`
`1/2 mv ^(2) = mgL + 1/2 (m ^(2) g ^(2))/( k)`
`therefore v ^(2) =2gL + (mg ^(2))/( k )`
`v = (2 gL + (mg ^(2))/( k )) ^(1//2)`
(c) When stone is at the lowest point Q, means at instantaneous distance Y from P from where the stone is dropped, then equation of motion of the stone is,
`(md ^(2)y )/( dt ^(2)) = mg - k (y -L)`
`implies (d ^(2) y )/( dt ^(2)) + k/m (y-L) -g =0`
Put another variable,
`z = k/m (y-L) -g`
`therefore ( d ^(2) z )/( dt ^(2)) + k/m z=0`
This is second dervative equation which indicates SHM.
Comparing this equation with
`(d ^(2) z )/(dt ^(2)) + omega ^(2) =0`
Angular frequency `omega = sqrt ((k)/(m))`
Equation simplify as under,
`z = A cos (omega t + phi),`
where `omega sqrt((k )/(m))`
`y = (L + (mg )/(k))+A. cos (omega t + phi)`
Hence, stone will execute SHM with angular frequency `omega = sqrt ((k)/(m))` about the point
`y _(0) = L+ (mg)/(k)`