A stone of mass m is tied to an elastic string of negligble 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

(a) As the stone falls vertically through length L, it is in free fall. After that the elasticity of the string will play its part and it will execute simple harmonic motion. Let the stone come to rest instantaneously at distance y from point P.

Ac cording to law of conservation of mechanical energy. the loss in potential energy of ston is equal to the elastic potential energy stored in a stetched string. Therefore, `mg y = (1)/(2) k (y -L)^2`
or `mg y =(1)/(2) k y^2 - k y L+(1)/(2) k L^2`
or `(1)/(2) k y^2 - (k L + mg) y + (1)/(2) k L^2 = 0`
`or y = (k L+mg) + -sqrt((k L+mg)^2 - k^2 L^2))/(k) = ( kL + mg)+ - sqrt(2mg k L+ m^2 g^2)/(k)`
Taking positive sign in consideration, we have `y = ((kL + mg) + sqrt(2mg k L+ m^2 g^2))/(k)`
(b) in SHM, the maximum velocity is attained when the stone passes through the equilibrium position. At this location, instanteous ac celeration is zero. If x is the extension in the string from length L at equilibrium position, then mg =kx or x = mg/k ... (i) Let `upsilon` be the velocity of stone at equilibrium position, then ac cording to law of conservation of energy `(1)/(2)m upsilon^2+(1)/(2) kx^2 = mg(L +x) or (1)/(2) m upsilon^2 = mg(L +x) - (1)/(2) kx^2`
or `(1)/(2)m upsilon^2 = mg( L+(mg)/(k)) - (1)/(2)k(m^2 g^2)/(k^2)` [from (i)]
`=mg L+(m^2 g^2)/(k) - (1m^2 g^2)/(k) = mg L+(m^2 g^2)/(2k)`
`or upsillon^2 = 2g L+ mg^2//k or upsilon = sqrt(2g L + mg^2//k)`
(c ) When stone is at the lowest position i.e., at instaneous distance y from P, then equation of motion of stone is `m(d^2y)/(dt^2) = mg -k (y -L) or m = (d^2y)/(dt^2) + k (y -L) - mg = 0`
`or (d^2y)/(dt) + (k)/(m) (y - L) - g = 0 .... (ii)`
Let `z = (k)/(m) (y -L) -g`
`:. (dz)/(dt) = (k)/(m)[(dy)/(dt) -0] -0 =(k)/(m) (dy)/(dt) or (d^2 z)/(dt^2) = (k)/(m) (d^2y)/(dt^2) or (d^2 y)/(dt^2) = (m)/(k) (d^2 z)/(dt^2)`
Putting this value in (ii), we get ` (m)/(k) (d^2 z)/(dt^2) +z = 0 or (d^2z)/(dt^2) +(k)/(m)z =0 ..... (iii)`
it is a differential equation of second order which represents SHM. Comparing (iii) with `(d^2 z)/(dt^2) + omega^2z =0.` we have Angular frequency of harmonic motion, `omega =sqrt((k)/(m))` The solution of (iii) will be of the type ` z = A cos (omega t + phi)`
or `y = L+(mg)/(k) +(mA)/(k) cos (omega + phi) = (L +(mg)/(k)) + A cos (omega t +phi) [where(mA)/(k) = A']`
Thus the stone will perform SHM with angular frequency `omega = sqrt(k//m)` about a point `y_0 = (L + mg//k).`