The periodic time of an elastic string's oscillation

[patric44]

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

 

[TSny]

$$T = 3mg$$

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?

 

[patric44]

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]

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).

 

[patric44]

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 :
$$ y′′(t)+\frac{g}{a}y(t)=g , y(0)=3a , y′(0)=0 $$
the answer was ?! :

what am i doing wrong ?

 

[haruspex]

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 :
$$ y′′(t)+\frac{g}{a}y(t)=g , y(0)=3a , y′(0)=0 $$
the answer was ?! :
View attachment 259376
what am i doing wrong ?

How else might you write e^iy?

 

[patric44]

How else might you write e^iy?

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]

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.

 

[patric44]

Close. Check to see if this equation is dimensionally correct.

You need to consider the possibility that the string becomes slack.

its not clear to me how its not dimensionally correct , inst the cos term has no units and a has a unit of length
same as y ?
- what you mean by slack ? and how should it effect my approach for solving the problem ?

 

[TSny]

its not clear to me how its not dimensionally correct , inst the cos term has no units and a has a unit of length
same as y ?

Consider $a + 2\cos(\sqrt{\frac{g}{a}}t)$

You are adding two terms. The first term is $a$ and has the dimensions of length. The second term is $2\cos(\sqrt{\frac{g}{a}}t)$ and is dimensionless. You can't add two terms of different dimensions.

- what you mean by slack? and how should it effect my approach for solving the problem ?

Try to picture what's going on. After releasing the mass, the mass will move upward. If it moves far enough upward, the string will become "loose" ("slack"). While the string is slack, it does not exert a force on the mass.

Some things for you to consider:

(1) For what value of $y$ does the string first become slack?

(2) While the string is slack, what force or forces act on the mass?

(3) How would you describe the motion of the mass while the string is slack?