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Extension of springs

Carla1985

Member
Feb 14, 2013
93
Second question im stuck on:

A spring of natural length l with modulus of elasticity λ has one end fixed to the ceiling. A particle of mass m is attached to the other end of the spring and is left to hang in its equilibrium position under the influence of gravity.

(i) Find the extension $x_E$ of the spring in the equilibrium position.

(ii) The particle is now lowered a distance $a$ below its equilibrium position and released from rest. Working from N2, show that if $z$ denotes the vertical displacement of the particle below its equilibrium position and we neglect theeffects of air resistance, then

$\ddot{z}+ω^2z = 0$,

where $ω^2 = λ/(lm)$. Deduce that $z(t) = a cos(ωt)$.

(iii) Suppose now that the particle is initially at the equilibrium position and is given a positive downward initial velocity $\dot{z}(0) = b$. Show that in this case,

$z(t) = \frac{b}{ω} sin ωt$,

and hence find the first time when the particle comes to rest."

I'm really stuck on this one. I know the position vector of the particle is $\vec{r}=(l+x)\vec{k}$

and I think $T=\frac{\lambda x_E}{l}$ so $x_E=\frac{Tlm}{\lambda}$ but not sure if thats way off.

Thanks
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,790
Second question im stuck on:

A spring of natural length l with modulus of elasticity λ has one end fixed to the ceiling. A particle of mass m is attached to the other end of the spring and is left to hang in its equilibrium position under the influence of gravity.

(i) Find the extension $x_E$ of the spring in the equilibrium position.

(ii) The particle is now lowered a distance $a$ below its equilibrium position and released from rest. Working from N2, show that if $z$ denotes the vertical displacement of the particle below its equilibrium position and we neglect theeffects of air resistance, then

$\ddot{z}+ω^2z = 0$,

where $ω^2 = λ/(lm)$. Deduce that $z(t) = a cos(ωt)$.

(iii) Suppose now that the particle is initially at the equilibrium position and is given a positive downward initial velocity $\dot{z}(0) = b$. Show that in this case,

$z(t) = \frac{b}{ω} sin ωt$,

and hence find the first time when the particle comes to rest."

I'm really stuck on this one. I know the position vector of the particle is $\vec{r}=(l+x)\vec{k}$

and I think $T=\frac{\lambda x_E}{l}$ so $x_E=\frac{Tlm}{\lambda}$ but not sure if thats way off.

Thanks
Hi again Carla! ;)

The modulus of elasticity λ is a property of a material.
To use it, you need the area of a cross section of the material.
So you need to use that
\(\displaystyle \qquad \displaystyle \lambda = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{x_E/l}\)
where F is the force of gravity and A is the area of a cross section.


Note that this can be rewritten as
\(\displaystyle \qquad F=\frac{\lambda A}{l} x_E = k x_E\)
which is the usual form of Hooke's law, meaning that the extension of a spring is linear with the applied force.


Btw, in the equilibrium position, you would have that $T$ is equal to $mg$.
 

Carla1985

Member
Feb 14, 2013
93
So the extension of the spring is $x_E=F\frac{l}{\lambda A}$? I'm a little confused by the A if I'm honest. In class we've just used λx/l and have never mentioned using the area of anything :/
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,790
So the extension of the spring is $x_E=F\frac{l}{\lambda A}$? I'm a little confused by the A if I'm honest. In class we've just used λx/l and have never mentioned using the area of anything :/
Yes.

It appears you are using a different modulus of elasticity in class.
My reference is the version on wiki: Elastic modulus or Young's modulus.

But if you're using F=λx/l in class, you'd get $x_E=F\frac{l}{\lambda}$.

Then it seems as if you've only made a mistake substituting $T=mg$.
 

Carla1985

Member
Feb 14, 2013
93
Yes, and the rest of the question makes a lot more sense now :) Thank you ever so much for your help. It is very much appreciated x