• Support PF! Buy your school textbooks, materials and every day products Here!

Transverse Oscillation

  • Thread starter e(ho0n3
  • Start date
  • #1
1,357
0
A mass m is connected to two springs with equal spring constants k. In the horizontal position shown, each spring is streched by an amount [itex]\Delta a[/itex]. The mass is raised vertically and begins to oscillate up and down. Assuming that the displacement is small, and ignoring gravity, show that the motion is simple harmonic and find the period.

Here is my analysis: The mass is streched vertically such that the springs strech an amount [itex]a + \Delta a[/itex]. The restoring force, F, is acting in the direction of the springs. The horizontal component of F from both springs cancel out but the vertical component from both springs are both in the same direction. Let [itex]F \sin \theta[/itex] be the vertical component of F, [itex]\theta[/itex] being the angle the string makes with the horizontal. The sum of the vertical forces acting on m, ignoring gravity is [itex]2 F \sin \theta[/itex].

What is [itex]\sin \theta[/itex]? Some quick trig. yields:

[tex]\sin \theta = \sqrt{(a + \Delta a)^2 - a^2} / (a + \Delta a) = \sqrt{2a \Delta a + (\Delta a)^2} / (a + \Delta a)[/tex]

Since the displacement is small, it is "safe" to approximate [itex]\sin \theta[/itex] as:

[tex]\sin \theta = \sqrt{2a \Delta a} / a = \sqrt{2 \Delta a / a} [/tex]

And so the sum of the vertical forces on m is:

[tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex]

The above is not akin to Hooke's law so I'm wondering how in the world is the motion simple harmonic? Perhaps I did something wrong in my analysis...
 
Last edited:

Answers and Replies

  • #2
960
0
A picture is worth a 1000 words; I'm assuming that you have:
spring--mass--spring under tension and horizontal, than displace the mass in the y direction, and it bobs up and down?
 
  • #3
960
0
if so, then the expression for force along the Y axis, Fy, is then:

-2k(sin(theta))*l where l=length of string/spring. This could be also written as
m*y"+2*k*y(t)=0.

This is a second order ODE that describes periodic motion.

Just like a bass guitar string. Intuitively I know that the original delta a will affect the tuning frequency, need a minute to figure out how mathematically..
 
Last edited:
  • #4
960
0
I believe the restoring force is 2sin(theta)*-2K*deltaA using the prestretched force figure and assuming small displacements about the equilibrium. In other words the result is the same as ypu have, this is still of the form,
my"+Ky=0 where (kY)>1, and so described undamped harmonic motion.
 
  • #5
1,357
0
A picture is worth a 1000 words; I'm assuming that you have:
spring--mass--spring under tension and horizontal, than displace the mass in the y direction, and it bobs up and down?
Yes, exactly. It's a shame that I don't have a scanner to reproduce the picture.

I believe the restoring force is 2sin(theta)*-2K*deltaA using the prestretched force figure and assuming small displacements about the equilibrium. In other words the result is the same as ypu have, this is still of the form,
my"+Ky=0 where (kY)>1, and so described undamped harmonic motion.
As I understand it, if the restoring force obeys Hooke's law, i.e. if the restoring force is proportional to the product of the displacement and some constant and the force points in the direction of the equilibrium point, then the motion is simple harmonic. The problem as I see it is that the constant is not a constant (it depends on the displacement).
 
  • #6
960
0
But look at hooke's law which if you were to simply have one spring attached and put the mass horizontally on an airtable and pull against the spring, it would be periodic as in f=-kx.

The eqn for position comes as a result of having conjugate imaginary roots e^ix and e-ix (the sqrty of x actually) which thru Eulers identity iirc, can be represented as sum of sin and cos, or lumped together in a single trig fx.
 
Last edited:
  • #7
1,357
0
But look at hooke's law which if you were to simply have one spring attached and put the mass horizontally on an airtable and pull against the spring, it would be periodic as in f=-kx.
That is my point. What is the force constant in

[tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex] ?

It is [itex] -2k \sqrt{2 \Delta a / a}[/itex] which is dependent on the displacement, i.e. it is not a constant.
 
Last edited:
  • #8
960
0
I see what you are asking now--the original displacement delta a. Sorry. If i understand the problem correctly, thats why the additional displacement is assumed small. So delta A is a "constant" in this case.
 
  • #9
1,357
0
I see what you are asking now--the original displacement delta a. Sorry. If i understand the problem correctly, thats why the additional displacement is assumed small. So delta A is a "constant" in this case.
That doesn't make sense. Just because the displacement is small, doesn't mean it is constant. If that were true, then the restoring force would be constant and there would be no SHM.
 
  • #10
960
0
No, not what I meant: the original stretch in line of springs, delta A, (the prestretch) is a constant, the transverse amplitude is both small and variable. Hope that helps.
 
  • #11
1,357
0
No, not what I meant: the original stretch in line of springs, delta A, (the prestretch) is a constant, the transverse amplitude is both small and variable. Hope that helps.
At an instance in time, any force is constant. If the amplitude is variable, i.e. it varies over time, then the motion is not simple harmonic. It may be damped harmonic motion or something else.

My point is that the equation I get does not look like [itex]F = -kx[/itex]. It looks like [itex]F = -k(x) x[/itex] where k(x) is a function of x, the displacement, or it looks like [itex]F = -k x^{3/2}[/itex]. Hence, it is NOT simple harmonic.
 
  • #12
960
0
I see it more like a pendulum with tension 2*T where T=-k*Delta(a), and the restoring force= sin (theta) * T =x/L * T
 
  • #13
1,357
0
By the way, the period according the book's answer is [itex]2 \pi (ma/2k \Delta a)^{1/2}[/itex], but according to my calculations (see post 7), it would be

[tex]2 \pi \sqrt{\frac{m}{ 2k \sqrt{2 \Delta a / a}}[/tex]

Because of the difference, I'm wondering if my analysis is wrong to begin with.
 
  • #14
960
0
what if we were to multiply both top and bottom of the quantity within the radical by a.
 
  • #15
Doc Al
Mentor
44,892
1,143
A mass m is connected to two springs with equal spring constants k. In the horizontal position shown, each spring is streched by an amount [itex]\Delta a[/itex]. The mass is raised vertically and begins to oscillate up and down. Assuming that the displacement is small, and ignoring gravity, show that the motion is simple harmonic and find the period.

Here is my analysis: The mass is streched vertically such that the springs strech an amount [itex]a + \Delta a[/itex]. The restoring force, F, is acting in the direction of the springs. The horizontal component of F from both springs cancel out but the vertical component from both springs are both in the same direction. Let [itex]F \sin \theta[/itex] be the vertical component of F, [itex]\theta[/itex] being the angle the string makes with the horizontal. The sum of the vertical forces acting on m, ignoring gravity is [itex]2 F \sin \theta[/itex].

What is [itex]\sin \theta[/itex]? Some quick trig. yields:

[tex]\sin \theta = \sqrt{(a + \Delta a)^2 - a^2} / (a + \Delta a) = \sqrt{2a \Delta a + (\Delta a)^2} / (a + \Delta a)[/tex]

Since the displacement is small, it is "safe" to approximate [itex]\sin \theta[/itex] as:

[tex]\sin \theta = \sqrt{2a \Delta a} / a = \sqrt{2 \Delta a / a} [/tex]

And so the sum of the vertical forces on m is:

[tex]2 F \sin \theta = 2 (-k \Delta a) \sqrt{2 \Delta a / a} [/tex]

The above is not akin to Hooke's law so I'm wondering how in the world is the motion simple harmonic? Perhaps I did something wrong in my analysis...
Here's how I would analyze this:

Since you are trying to express the motion in terms of the vertical displacement, I will assume that [itex]\Delta a[/itex] is the vertical displacement of the mass. When the mass is displaced that amount, the springs are stretched to a distance: [itex]\sqrt{a^2 + (\Delta a)^2}[/itex].

So the tension in each spring is now:

[tex]T = k \sqrt{a^2 + (\Delta a)^2}[/tex]

Thus the restoring force (the sum of the vertical components of tension from each spring) is now:

[tex]F = 2 k \sqrt{a^2 + (\Delta a)^2}\frac{\Delta a}{\sqrt{a^2 + (\Delta a)^2}}[/tex]

Which is just:

[tex]F = 2 k \Delta a[/tex]

The motion is clearly simple harmonic. (And it should make intuitive sense that the spring constant doubled, since you have two springs in parallel.)

By the way, the period according the book's answer is [itex]2 \pi (ma/2k \Delta a)^{1/2}[/itex]...
I don't understand this answer, since it implies that the period depends on the amplitude--that's not SHM. Seems like an error to me.
 
Last edited:
  • #16
1,357
0
Since you are trying to express the motion in terms of the vertical displacement, I will assume that [itex]\Delta a[/itex] is the vertical displacement of the mass. When the mass is displaced that amount, the springs are stretched to a distance: [itex]\sqrt{a^2 + (\Delta a)^2}[/itex].
I just dawned on me this morning. The displacement [itex]\Delta a[/itex] in my analysis is not the vertical displacement. The vertical displacement is [itex]\sqrt{(a + \Delta a)^2 - a^2}[/itex]. Let y be this vertical displacement. Then,

[tex]\sin \theta = \frac{y}{a + \Delta a}[/tex]

and so

[tex]2F \sin \theta = 2 (-k \Delta a) \frac{y}{a + \Delta a} = - \frac{2ky \Delta a}{a + \Delta a} [/tex].

For small [itex]\Delta a[/itex], the above is approximated by [itex]- (2k \Delta a / a)y[/itex]. This is akin to Hooke's law and so the motion is SHM. The period is

[tex] 2 \pi \sqrt{\frac{m}{2k \Delta a / a}} = 2 \pi \sqrt{\frac{ma}{2k \Delta a}}[/tex]

which does look like the books answer.
 
Last edited:
  • #17
960
0
nice job there.
 
  • #18
Doc Al
Mentor
44,892
1,143
This still doesn't seem quite right to me. You are using [itex]a[/itex] to represent the amount that the springs are stretched when they are in there equilibrium position, and [itex]a + \Delta a[/itex] is the amount that the springs are stretched when the mass is displaced vertically. (Note: The mass moves vertically. If the mass exhibits SHM, the restoring force must be proportional to its vertical displacement from equilibrium.)
I just dawned on me this morning. The displacement [itex]\Delta a[/itex] in my analysis is not the vertical displacement. The vertical displacement is [itex]\sqrt{(a + \Delta a)^2 - a^2}[/itex]. Let y be this vertical displacement. Then,

[tex]\sin \theta = \frac{y}{a + \Delta a}[/tex]
Looks good.

and so

[tex]2F \sin \theta = 2 (-k \Delta a) \frac{y}{a + \Delta a} = - \frac{2ky \Delta a}{a + \Delta a} [/tex].
I would say that the spring tension is given by [itex]k (a + \Delta a)[/itex], not [itex]k \Delta a[/itex], so that:

[tex]2F \sin \theta = -2 k (a + \Delta a) \frac{y}{a + \Delta a} = - 2 k y[/tex]

That is SHM about the equilibrium position.

For small [itex]\Delta a[/itex], the above is approximated by [itex]- (2k \Delta a / a)y[/itex]. This is akin to Hooke's law and so the motion is SHM.
Since [itex]\Delta a[/itex] and [itex]y[/itex] are not independent variables, this does not look like Hooke's law or SHM to me.

The period is

[tex] 2 \pi \sqrt{\frac{m}{2k \Delta a / a}} = 2 \pi \sqrt{\frac{ma}{2k \Delta a}}[/tex]

which does look like the books answer.
One of the characteristics of SHM is that the period is independent of amplitude, which is not the case with the book answer. So I don't see how this answer can be considered SHM. (If I'm missing something obvious, let me know! :cool: )

What text are you using?
 
  • #19
960
0
ignore, this/
 
Last edited:
  • #20
1,357
0
Since [itex]\Delta a[/itex] and [itex]y[/itex] are not independent variables, this does not look like Hooke's law or SHM to me.
You're right about that. Hooke's law requires a force constant and since [itex]\Delta a[/itex] is not constant, it does not satisfy Hooke's law.

One of the characteristics of SHM is that the period is independent of amplitude, which is not the case with the book answer. So I don't see how this answer can be considered SHM.
What text are you using?
I'm using "Physics for Scientists and Engineers with Modern Physics" by Douglas C. Giancoli: http://cwx.prenhall.com/bookbind/pubbooks/giancoli3/ [Broken]
 
Last edited by a moderator:
  • #21
1,357
0
I would say that the spring tension is given by [itex]k (a + \Delta a)[/itex], not [itex]k \Delta a[/itex]
Why would you say that? From what I understood, the mass/spring is in equilibrium when the spring has length a.
 
  • #22
960
0
well thats part of the confusion I have felt from the beginning, "a" was never really defined. delta a, was clear: the stretch exerted initially. I still think its shm, but purely from physical intuition at this point as being analogous to a pendulum under the small angle constraints. Or by resolving the spring forces at max displacement into a pair of horizontal springs whose effect is cancelled out and a vertical spring, which represents the restoring force along the y axis. I also think the energy constancy condition would be fulfilled. Not sure why this is such a conundrum as I agree with both OP and Doc Al that the periods looks fishy even if it satisfies dim analysis. Still wish i had a pic, any chance e(hoOn3) you can scan the problem?
 
  • #23
Doc Al
Mentor
44,892
1,143
I would say that the spring tension is given by [itex]k (a + \Delta a)[/itex], not [itex]k \Delta a[/itex]
Why would you say that? From what I understood, the mass/spring is in equilibrium when the spring has length a.
Just because the system is in equilibrium does not mean that the springs are not under tension.
well thats part of the confusion I have felt from the beginning, "a" was never really defined. delta a, was clear: the stretch exerted initially.
Good point. I take [itex]a[/itex] to be the length of the springs in their horizontal (equilibrium) position and [itex]\Delta a[/itex] to be their length when the mass is displaced from equilibrium.

In lieu of a picture, here's how I would describe the problem: A mass is suspended horizontally between fixed supports by means of two identical springs. Each spring has length [itex]a[/itex] when the mass is at the equilibrium position. The mass is displaced vertically so that the springs have length [itex]a + \Delta a[/itex]. I hope that's what we've all been talking about! :wink:

To get a more accurate answer, we should compare [itex]a[/itex] with [itex]a_0[/itex], the unstretched length of the springs. The restoring force is then:

[tex]2F \sin \theta = -2 k (a + \Delta a - a_0) \frac{y}{a + \Delta a}[/tex]

In the small displacement limit (when [itex]\Delta a << a[/itex]) this is:

[tex]2F \sin \theta = -2 k (\frac{a - a_0}{a}) y[/tex]

That's still SHM, but the effective spring constant is not quite 2k. In the limit where the initial tension is great ([itex]a >> a_0[/itex]), then the effective spring constant becomes just 2k.
 
Last edited:
  • #24
1,357
0
In lieu of a picture, here's how I would describe the problem: A mass is suspended horizontally between fixed supports by means of two identical springs. Each spring has length [itex]a[/itex] when the mass is at the equilibrium position. The mass is displaced vertically so that the springs have length [itex]a + \Delta a[/itex]. I hope that's what we've all been talking about!
Yes, that is the situation I was analyzing.

To get a more accurate answer, we should compare [itex]a[/itex] with [itex]a_0[/itex], the unstretched length of the springs.
From what I understood of the problem, a is the length of the unstreched spring. That's why the restoring force is proportional to [itex]\Delta a[/itex] and not [itex]a + \Delta a[/itex].
 
  • #25
960
0
thanks for the clarification. I am for one satisfied that the analysis in post 16 provides an effective k' for the problem which assumes delta A is an initial condition, a is constant, and that
A+delta A >>y.
 

Related Threads on Transverse Oscillation

Replies
5
Views
793
  • Last Post
Replies
3
Views
5K
Replies
13
Views
1K
Replies
2
Views
6K
Replies
14
Views
6K
  • Last Post
Replies
2
Views
10K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
6
Views
4K
Top