Understanding transverse oscillation in 1 mass, 2 spring systems

Question

Lately I have been working through some nice problems on mass-spring systems. There are tons of different configurations - multiple masses, multiple springs, parallel/series, etc.

A few possible configurations are listed here: https://ccrma.stanford.edu/CCRMA/Courses/152/vibrating_systems.html. I think the most interesting one is the 1 mass, 2 spring system with transverse motion:

It is written on that page:

"If the springs are initially stretched a great deal from their relaxed length (but not distorted), the vibration frequency is nearly the same as for longitudinal vibrations.

If the springs are initially stretched very little from their relaxed length, the "natural" frequency is much lower and the vibrations are nonlinear (nonsinusoidal) for all but the smallest of $y$-axis displacements."

If I understand correctly, this means that the system approaches simple harmonic motion when the initial displacement $y_0$ is either very small, or very large.

I tried to see if I could prove this myself by looking at the period of oscillations as a function of the initial displacement $y_0$, but I am having some trouble:

Suppose the mass is a distance $\Delta y$ from the equilibrium position. Then each spring has length $\sqrt {a^2+\Delta y^2}$, so each spring is stretched by $\sqrt {a^2+\Delta y^2}-a$ from its original length. Therefore, each spring exerts a restoring force of $k(\sqrt {a^2+\Delta y^2}-a)$. The magnitude of the net restoring force on the mass is then

$|F_r|=2k(\sqrt {a^2+\Delta y^2}-a)\sin(\theta)=2k(\Delta y-a\sin(\theta))$

Substituting $\sin(\theta)=\Delta y/\sqrt {a^2+\Delta y^2}$, we get:

$|F_r|=2k\Delta y(1-a/\sqrt {a^2+\Delta y^2})$

So we have the differential equation:

$y''=-(2k/m)y(1-a/\sqrt {a^2+y^2})$.

If $y$ is very large, then $y''\approx -(2k/m)y$, which is simple harmonic motion with period $2\pi\sqrt{2k/m}$. This makes sense, because if $y$ is very large, the two springs essentially act in parallel, so we effectively have a 1 spring system in longitudinal motion with spring constant $2k$, which gives the same result.

Now if $y$ is very small, I'm not sure which approximation formulas to use to make things come out nicely. According to the web page, I should get the result that the period will be larger than $2\pi\sqrt{2k/m}$.

Answer 1

I think you are misunderstanding what they mean by

If the springs are initially stretched very little from their relaxed length

You take it to mean "large displacement from equilibrium" - but I think it means "both springs are under tension in equilibrium". In that case there is a significant tension $T$ without displacement, and for small displacements $y$ the restoring force is approximately given by

$$F = T \sin\theta \approx T ~\frac{y}{a}$$

Because $T$ will not change much if there was quite a bit of initial tension, the system will be linear.

However, if there is little initial tension, then the restoring force will be mostly due to the additional stretching of the spring. In the case of zero initial tension, the restoring force (for small displacements) is given by

$$\begin{align}\\ F &= k\left(\sqrt{y^2+a^2}-a\right)\sin\theta\\ &=ka\left(\sqrt{1+\frac{y^2}{a^2}}-1\right)\frac{y}{a}\\ &\approx\frac{ky^3}{2a^2} \end{align}$$

So when the increase in tension due to the lateral displacement is significant, the motion becomes non-linear.

UPDATE

We can look at this for both horizontal and vertical displacements. Assume the unstretched length of the springs is $L_0$, and the stretched length (in equilibrium) is $L\gg L_0$. If we displace by a small amount $dx$ horizontally, and a small amount $dy$ vertically (both $\ll LL$), then we can compute the horizontal and vertical forces on the mass.

First we compute the new length of the spring. This will be

$$L_1 = \sqrt{(L+dx)^2+dy^2}$$

for the spring on the left, and

$$L_2 = \sqrt{(L-dx)^2+dy^2}$$

for the spring on the right.

The net force on the mass is given by the horizontal and vertical components of the tension in the spring.

For the horizontal component, we notice that

$$F_x = T_1 \cos\alpha_1 - T_2\cos\alpha_2$$

for small angles this simplifies to

$$F_x = T_1 - T_2\\ = kL\left[\left(\left((1+\frac{dx}{L}\right)^2+\left(\frac{dy}{L}\right)^2\right)^{1/2}-\left(\left((1-\frac{dx}{L}\right)^2+\left(\frac{dy}{L}\right)^2\right)^{1/2}\right]\\ \approx 2kdx$$

For the vertical component of force, we approximate the sum of $T_1$ and $T_2$, and find that it varies only with higher orders of $dx$ and $dy$ - so we consider it ($T_1+T_2$) constant for small displacements. The vertical force $F_y = 2T\sin\alpha\approx 2T\frac{dy}{L}$ will then depend only on $dy$ and not on $dx$. So horizontal and vertical oscillations will be independent, and linear.

Answer 2

Please see the lecture on http://www.unizor.com under Physics 4 Teens > Waves > Transverse Waves > Musical Strings 1. It models a musical string (like the one on a violin) with a model you describe. It explains why oscillations with small vertical deviation are close to harmonic. The angular frequency $\omega$ of these oscillations is $$\omega^2 = \frac{2κ}{[m(1+L/l)]}$$ where
$k$ - elasticity of a string,
$m$ - mass,
$L$ - length in neutral state,
$l$ - initial stretch to create tension.

In particular, it's seen from this formula that the higher the initial tension (higher initial stretch $l$), the higher the angular frequency $\omega$, that is, the higher tone of a sound produced by a string.