# Oscillating rod attached to spring

## Homework Statement

A thin uniform rod is pivoted about its mid-point, with the lower end attached to a spring. See the figure. At equilibrium the rod is vertical. Show that on a (small) displacement from equilibrium the rod will oscillate with simple harmonic motion at a frequency:
$$f = \frac{1}{2 \pi} \sqrt{\frac{3k}{M}}$$
Where $k$ is the spring constant and $M$ is the mass of the rod.

## Homework Equations

$$F = ma\\ \tau = I \alpha\\ I = \frac{1}{12} ML^{2}\\ f = \frac{\omega}{2 \pi}\\ \sin(\theta) \approx \theta$$
I believe this is the moment of inertia for a uniform rod rotating about its mid-point

## The Attempt at a Solution

I tried to relate the displacement $x$ to the displacement $\theta$ using geometry and then write a well known differential equation with a well known solution to get $\omega$ and then do some massaging to get $f$. However my answer doesn't match.
$$\tau = I \alpha \\ -kx = \frac{1}{12}ML^{2} \alpha \\ \sin(\theta) = \frac{2x}{L}\\ x = \frac{L\theta}{2}\\ -k \frac{L\theta}{2} = \frac{1}{12}ML^{2} \ddot{\theta}\\ 0 = \frac{1}{6}ML\ddot{\theta} + k\theta\\ \omega = \left(\frac{6k}{ML}\right)^{\frac{1}{2}}\\ f = \frac{1}{2\pi}\left(\frac{6k}{ML}\right)^{\frac{1}{2}}\\$$

From the Euler-Lagrange equations ##\ddot{\theta}=-\frac{k'}{I}\theta##, so your equation (6) is correct with ##k'=kL/2##. The units given in the answer are different from yours ( with which I agree). Maybe that's a clue ?

• 1 person
TSny
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$$\tau = I \alpha \\ -kx = \frac{1}{12}ML^{2} \alpha \\$$

Is ##-kx## a torque or a force?

Is ##-kx## a torque or a force?
Yes, I think that's the problem. That equation should read ##-k'\theta = \frac{1}{12}ML^{2} \alpha##.

TSny
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I think mbigras just needs to multiply the force (##-kx##) by the lever arm to get the torque. Then he can proceed as he did in his original post.

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Thank you all,

$$\Sigma \tau = I \ddot \theta\\ -kx\frac{L}{2} = \frac{1}{12} M L \ddot \theta\\ -k\theta \frac{L^{2}}{4} = \frac{1}{2} M L^{2} \ddot \theta\\ 0 = \frac{1}{3} M \ddot \theta + k \theta\\ f = \frac{1}{2\pi} \left(\frac{3M}{M}\right)^{\frac{1}{2}}$$

I think mbigras just needs to multiply the force (##-kx##) by the lever arm to get the torque. Then he can proceed as he did in his original post.

Sir ,

Please help me understand just one thing . Why do we take extension in the string to be LΘ/2 as the rod rotates by small angle Θ ? The end point of rod and hence the end of the spring attached move in an arc of length LΘ/2 . The extension in the spring is not equal to the distance moved by the tip of the rod . Then why do we assume that spring remains horizontal and is equal to the distance moved by the tip of the rod ?

Thanks

TSny
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Hello, Vibhor.

You are right that the spring does not remain exactly horizontal. However, you can show that the vertical displacement of the lower end of the rod is zero to first order in θ.

As an exercise, show that the horizontal and vertical displacements of the lower end of the rod are
Δx = (L/2)sinθ
Δy = (L/2)(1-cosθ).

What are the lowest order nonzero approximations of these expression for small θ?

• Vibhor
As an exercise, show that the horizontal and vertical displacements of the lower end of the rod are
Δx = (L/2)sinθ
Δy = (L/2)(1-cosθ).

What are the lowest order nonzero approximations of these expression for small θ?

Sinθ ≈ θ and Cosθ ≈ 1 . Using these Δx ≈ Lθ/2 and Δy ≈ 0 .

TSny
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Yes. If you expand cosθ to second order, you get that 1 - cosθ ≅ θ2/2. So, Δy is "second order" in θ and can be neglected compared to Δx for small θ.

• Vibhor
OK . But why should we compare Δy to Δx ? Aren't the displacements along x-y directions independent of each other ?

TSny
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We are making a "small angle approximation". For small θ, Δy is negligible compared to Δx. So, you can neglect Δy and tread the displacement of the end of the rod as being horizontal of amount Δx.

• Vibhor
Suppose the rod is hinged at the top instead of in the middle as in the OP . And we are asked to use Energy method to find the frequency . The energy at any θ would be

E = (1/2)Iω2 + (1/2)k(Lθ)2 + mgL(1-cosθ)/2 .

Why do we consider potential energy of the rod taking into account vertical displacement of the CM of the rod (last term )And at the same time neglect the vertical displacement of the tip attached to the spring(middle term) ?

TSny
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Suppose the rod is hinged at the top instead of in the middle as in the OP . And we are asked to use Energy method to find the frequency . The energy at any θ would be

E = (1/2)Iω2 + (1/2)k(Lθ)2 + mgL(1-cosθ)/2 .

Why do we consider potential energy of the rod taking into account vertical displacement of the CM of the rod (last term )And at the same time neglect the vertical displacement of the tip attached to the spring(middle term) ?
The lowest order nonzero terms in the energy are second-order terms in θ or ω. So, each of the three terms should be written to second order when considering small oscillations. The first term is already of second-order. The last term will be of second-order if you expand 1 - cosθ to order θ2. The middle term is accurate to second order if you treat Δx in the stretch of the spring to first order in θ. Including Δy in the stretch of the spring would not include any first or second order contribution to the middle term. The effect of Δy would be to introduce terms of higher than second order to the middle term. But these higher order terms are neglected in the case of small oscillations.

• Vibhor
Including Δy in the stretch of the spring would not include any first or second order contribution to the middle term. The effect of Δy would be to introduce terms of higher than second order to the middle term.

Last edited:
TSny
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The potential energy of the spring is proportional to the square of the amount of stretch of the spring. You only need to find the stretch of the spring to first-order in θ. Then the potential energy will be correct to second order. Δy does not have a first-order contribution to the stretch of the spring.

To really see this, you can first derive an expression for the total length of the spring as a function of θ for arbitrary θ (not necessarily small). This expression will contain sinθ (from Δx) and cosθ (from Δy). Then, Taylor expand the expression about θ = 0 and keep only terms up to first order in θ. You will see that there is no first-order contribution from Δy.

• Vibhor
It should be
-KxL/2=ML²α/12​

TSny
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It should be
-KxL/2=ML²α/12​
Yes, that's correct. In post #6 @mbigras wrote
$$\Sigma \tau = I \ddot \theta\\ -kx\frac{L}{2} = \frac{1}{12} M L \ddot \theta\\ -k\theta \frac{L^{2}}{4} = \frac{1}{2} M L^{2} \ddot \theta\\ 0 = \frac{1}{3} M \ddot \theta + k \theta\\ f = \frac{1}{2\pi} \left(\frac{3M}{M}\right)^{\frac{1}{2}}$$
The second equation has a typo in that the ##L## should be squared on the right side of the equation. He/she fixed this in the third equation. But unfortunately a new typo occurs in the third equation where the ##\frac{1}{2}## on the right should be ##\frac{1}{12}##. The fourth equation is correct. The last equation has a typo where the ##M## in the numerator of the fraction should be ##k##.