# The periodic time of an elastic string's oscillation

## 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 :
$$Tension = \frac {λ*y}{a}$$
$$\lambda= mg$$
$$y =3a$$
$$T = 3mg$$
$$F = T-mg\Longrightarrow F = 3mg-mg = 2mg$$
$$m{y}''=2mg$$
$$y'' = 2g \therefore\frac { dy'}{dt} = 2g \Longrightarrow y' = 2gt+c1$$ by applying the boundary conditions and integrating
$$y = gt^2 -3a$$
$$t^2=\frac{y-3a}{g}⇒ t =√(\frac{a}{g})√(\frac{y}{a} +3)$$
i am stuck here what should i do next
is this the right approach to solve it ?

i will appreciate any help
thanks

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TSny
Homework Helper
Gold Member
$$T = 3mg$$
This is true only at the instant where $y=3a$.
$$F = T-mg\Longrightarrow F = 3mg-mg = 2mg$$
$$m{y}''=2mg$$
This is true only when $y = 3a$.
You will need expressions for $F$ and $y''$ that hold for any positive value of $y$. Does $y$ ever go negative?

This is true only at the instant where $y=3a$.
This is true only when $y = 3a$.
You will need expressions for $F$ and $y''$ that hold for any positive value of $y$. Does $y$ ever go negative?
So I should work this with y as a variable, bud this led me to some very lengthy calculations :

TSny
Homework Helper
Gold Member
I won't be able to easily quote particular parts of your work since it is posted as a picture.

Be careful with signs. Your diagram shows that you are taking positive $y$ downward. That's fine. But then, the weight of the mass $mg$ would be a positive force while the tension would be negative. So, your signs don't look right when you wrote $F = T - mg$.

For a mass hanging on a spring or elastic string, it is often a good idea to introduce a variable that measures the displacement of the mass from the equilibrium position (where the mass would hang at rest). You should find that you get a familiar differential equation that will yield a familiar type of motion (as long as the string doesn't go slack).

I won't be able to easily quote particular parts of your work since it is posted as a picture.

Be careful with signs. Your diagram shows that you are taking positive $y$ downward. That's fine. But then, the weight of the mass $mg$ would be a positive force while the tension would be negative. So, your signs don't look right when you wrote $F = T - mg$.

For a mass hanging on a spring or elastic string, it is often a good idea to introduce a variable that measures the displacement of the mass from the equilibrium position (where the mass would hang at rest). You should find that you get a familiar differential equation that will yield a familiar type of motion (as long as the string doesn't go slack).
so you are suggesting that the differential equation would be :
$$my''=mg-\frac{λ}{a}y ⇒y''=g-\frac{g}{a}y$$
is that the right differential equation to solve ?
i tried to solve it using online calculator to save some time :
$$y′′(t)+\frac{g}{a}y(t)=g , y(0)=3a , y′(0)=0$$

what am i doing wrong ?

haruspex
Homework Helper
Gold Member
so you are suggesting that the differential equation would be :
$$my''=mg-\frac{λ}{a}y ⇒y''=g-\frac{g}{a}y$$
is that the right differential equation to solve ?
i tried to solve it using online calculator to save some time :
$$y′′(t)+\frac{g}{a}y(t)=g , y(0)=3a , y′(0)=0$$
View attachment 259376
what am i doing wrong ?
How else might you write eiy?

How else might you write eiy?
i could write this as :
after expressing this in euler's form and canceling the sin :
$$y(t) = a+2\cos( {\sqrt{\frac{g}{a}}t})$$
so
$$t = \sqrt{\frac{a}{g}}\cos^{-1} ({\frac{y-a}{2}})$$
how would i get the periodic time as :
$$t = \sqrt{\frac{a}{g}}*(2√ 3 + 4π/3 )$$

TSny
Homework Helper
Gold Member
i could write this as :
after expressing this in euler's form and canceling the sin :
$$y(t) = a+2\cos( {\sqrt{\frac{g}{a}}t})$$
Close. Check to see if this equation is dimensionally correct.

how would i get the periodic time as :
$$t = \sqrt{\frac{a}{g}}*(2√ 3 + 4π/3 )$$
You need to consider the possibility that the string becomes slack.