# Math Help - 2 elastic light springs

1. ## 2 elastic light springs

Hi
I cant seem to get a diagram for this question. Can anyone help and also please explain the subsequent diagram. This is the question. A particle of mass 0.8kg is attached to the ends of 2 identical light elastic springs of natural length 1.6m and modulus of elasticity 16N. The free ends of the springs are attached to 2 points where points A and B which are 4m apart on a smooth horizontal surface. The point C lies between A and B such that ABC is a straight line and AC = 2.5m. The particle is held at C and then released from rest.
My textbook has tensions acting in different directions. Can anyone explain? Thanks.

2. Originally Posted by gbenguse78
Hi
I cant seem to get a diagram for this question. Can anyone help and also please explain the subsequent diagram. This is the question. A particle of mass 0.8kg is attached to the ends of 2 identical light elastic springs of natural length 1.6m and modulus of elasticity 16N. The free ends of the springs are attached to 2 points where points A and B which are 4m apart on a smooth horizontal surface. The point C lies between A and B such that ABC is a straight line and AC = 2.5m. The particle is held at C and then released from rest.
My textbook has tensions acting in different directions. Can anyone explain? Thanks.
The force exerted by a spring is opposite the direction of the spring's displacement from equilibrium. From your description, the spring attached to A is stretched 0.9 m while the spring attached to B is compressed 0.1 m. The forces acting on the mass are shown in my rudimentary diagram (I failed crayola manipulation in kindergarten ... not much better with a mouse.) ... so, I disagree w/ your book.

3. Originally Posted by skeeter
The force exerted by a spring is opposite the direction of the spring's displacement from equilibrium. From your description, the spring attached to A is stretched 0.9 m while the spring attached to B is compressed 0.1 m. The forces acting on the mass are shown in my rudimentary diagram (I failed crayola manipulation in kindergarten ... not much better with a mouse.) ... so, I disagree w/ your book.
Thats very clear. So that means using Hookes Law:
Tension in A =(16X0.9)/1.6
Tension in B =(16X0.1)/1.6

The rest of the question goes like this:
(a) Show that the subsequent motion is simple harmonic motion.
(b) Find the period and amplitude of the motion.
(c)Calculate the maximum speed of the particle.

Solution a
Is this correct
Tension in B-Tension in A=0.8 X the acceleration?

4. Yes, that is correct.

It is interesting that, although you are given the 'natural length' of the two springs, you don't need it.

Suppose we have two springs, each with modulus of elasticity k N/m (NOT Newtons) and natural length l. The two springs are attached to points distance L m apart and to an object of mass m. Let x be the distance from the object to the "left" point of attachment. Then the spring on the left has length x and is extended to distance x- l beyond its natural length and so applies force -k(x- l) to the object. The spriing on the right has length L- x and so is extended L- x- l beyond its natural length and applies force k(L- x- l) to the object. The total force on the object is -k(x- l)+ k(L- x- l)= -kx+ kl+ kL- kx- kl= kL- 2kx.

"Force= mass times acceleration" gives, as the "equation of motion",
$m\frac{d^2x}{dt^2}= kL- 2kx$
so the motion is independent of the natural length "l".

5. Originally Posted by HallsofIvy
Yes, that is correct.

It is interesting that, although you are given the 'natural length' of the two springs, you don't need it.

Suppose we have two springs, each with modulus of elasticity k N/m (NOT Newtons) and natural length l. The two springs are attached to points distance L m apart and two an object of mass m. Let x be the distance from the object to the "left" point of attachment. Then the spring on the left has length x and is extended to distance x- l beyond its natural length and so applies force -k(x- l) to the object. The spriing on the right has length L- x and so is extended L- x- l beyond its natural length and applies force k(L- x- l) to the object. The total force on the object is -k(x- l)+ k(L- x- l)= -kx+ kl+ kL- kx- kl= kL- 2kx.

"Force= mass times acceleration" gives, as the "equation of motion",
$m\frac{d^2x}{dt^2}= kL- 2kx$
so the motion is independent of the natural length "l".
Thanks. Really appreciated.