Please help in solving this question ?

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1 Answer
Sep 23, 2017

See below.

Explanation:

At the instant of release

#1/2kL^2=1/2(m+M)v_0^2# (Energy)

here #v_0# is the initial velocity of #m+M#

after release

#-k(x-x_0) = (m+M) ddot x# (Newtonian mechanics)

and solving

#x = x_0 + C_1 cos(omegat)+C_2 sin(omegat)#

with #omega = sqrt(k/(m-M))#

The determination of #C_1,C_2# is made according to the initial conditions

#{(x(0)=x_0-L),(dot x(0) = v_0):}#

giving

#x = x_0 - L cos(omega t) + v_0/omega sin(omega t)#

Now, #m,M# trip together until #ddot x = 0# or when

#omega (L omega Cos(omega t) - v_0 Sin(omega t)) = 0# at

#t = arctan(v_0/sqrt[L^2 omega^2 + v_0^2], (L omega)/sqrt[ L^2 omega^2 + v_0^2])/omega#

The last item is left as an exercise.

NOTE:

Integration of #-k(x-x_0) = (m+M) ddot x#

First making #y = x-x_0# we have

#-k/(m+M)y=ddot y# now assuming #y = e^(lambda t)# and substituting

#(k/(m+M)+lambda^2)e^(lambda t)=0# so #lambda = pm iomega# where

#omega = sqrt(k/(m+M))# then

#y = c_1 e^(i omega t)+c_2 e^(-iomega t)#

now using the de Moivre's identity

#e^(i omega t) = cos(omegat)+isin(omegat)# after some algebric considerations we get equivalently

#y = C_1 cos(omegat)+C_2sin(omegat)# (note that #c_i ne C_i#)

and finally

#x = x_0 +C_1 cos(omegat)+C_2sin(omegat)#