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Homework Statement:

a string with length "a" attached to a point and has a mass "m " attached to it , the elasticity of the string λ = the weight of that mass . if the mass is pulled downwards such that the string length becomes 4 times it original length . and its left to oscillate .
prove that mass will return back to that point of pulling for the first time after a time = √ (a/g)*(2√ 3 + 4π/3 )
Relevant Equations:

prove that t = √ (a/g)*(2√ 3 + 4π/3 )
λ =mg
i guess he is asking for the periodic time :
$$Tension = \frac {λ*y}{a} $$
$$
\lambda= mg
$$
$$y =3a$$
$$T = 3mg$$
$$F = Tmg\Longrightarrow F = 3mgmg = 2mg$$
$$m{y}''=2mg$$
$$y'' = 2g \therefore\frac { dy'}{dt} = 2g \Longrightarrow y' = 2gt+c1$$ by applying the boundary conditions and integrating
$$y = gt^2 3a $$
$$ t^2=\frac{y3a}{g}⇒ t =√(\frac{a}{g})√(\frac{y}{a} +3) $$
i am stuck here what should i do next
is this the right approach to solve it ?
i will appreciate any help
thanks
$$Tension = \frac {λ*y}{a} $$
$$
\lambda= mg
$$
$$y =3a$$
$$T = 3mg$$
$$F = Tmg\Longrightarrow F = 3mgmg = 2mg$$
$$m{y}''=2mg$$
$$y'' = 2g \therefore\frac { dy'}{dt} = 2g \Longrightarrow y' = 2gt+c1$$ by applying the boundary conditions and integrating
$$y = gt^2 3a $$
$$ t^2=\frac{y3a}{g}⇒ t =√(\frac{a}{g})√(\frac{y}{a} +3) $$
i am stuck here what should i do next
is this the right approach to solve it ?
i will appreciate any help
thanks