# Spring attached to a pendulum

• Ujjwal Basumatary
In summary, a pendulum of mass ##m## and length ##L## is connected to a spring as shown in figure. If the bob is displaced slightly from its mean position and released, it performs simple harmonic motion. The angular frequency of the bob is given by $$\omega=\sqrt{\frac{\alpha}{\theta}}=\sqrt{\frac{K}{I}}$$ where ##\theta=## Displaced angle, ##\alpha=## Angular acceleration, ##I=## Moment of inertia, and ##K=## Equivalent force constant.f

## Homework Statement

A pendulum of mass ##m## and length ##L## is connected to a spring as shown in figure. If the bob is displaced slightly from its mean position and released, it performs simple harmonic motion. What is the angular frequency of the bob?

## Homework Equations

Angular frequency for angular simple harmonic motion is given by $$\omega=\sqrt{\frac{\alpha}{\theta}}=\sqrt{\frac{K}{I}}$$
Where
##\theta=## Displaced angle
##\alpha=## Angular acceleration
##I=## Moment of inertia
##K=## Equivalent force constant

## The Attempt at a Solution

I have done the problem in the following way:
Consider a small angular displacement of ##\theta##. The torque about the the hinge at any instant is then given by $$\Gamma = -mgL\sin\theta-kxh$$ where ##x## is the displacement of the spring. But ##x=h\sin\theta## and therefore our equation becomes $$\Gamma = -mgL\sin\theta -mh^2\sin\theta$$
But ##\sin\theta \approx \theta## for small ##theta##. Therefore the torque is given by $$\Gamma = I\alpha=-mgL\theta -mh^2\theta$$ Thus we have ##\Gamma \propto -\theta##, which is the condition for angular SHM. Recalling that ##I=mL^2## we finally obtain $$\omega=\sqrt{\frac{\alpha}{\theta}} = \sqrt{\frac{K}{I}} = \sqrt{\frac{kh^2+mgL}{mL^2}}$$

Please look and tell me if there's anything wrong with the solution. Thank you. #### Attachments

Hello. Your work looks correct to me.

• Delta2
Is the bob suspended by a flexible string or a rigid rod?

tell me if there's anything wrong with the solution.
In a couple of places you have mh2 instead of kh2, but that just looks like a typo in writing the post.

Is the bob suspended by a flexible string or a rigid rod?
That's interesting, since this is classified as introductory physics homework, I don't think it can be a flexible string. How would you treat the problem to find the waves on the string if we had a flexible string?

If the string is massless there would be no waves. But there would be two different angles of string to deal with.
https://en.wikipedia.org/wiki/Complex_harmonic_motion#Double_Pendulum

Hmmmm double pendulum has two mass weights, not sure if the radial component of the weight of the inbetween mass plays a key role. I won't lie I haven't study the double pendulum equations at all. My intuition says that it plays an important role cause it keeps the string stressed.

Hmmmm double pendulum has two mass weights, not sure if the radial component of the weight of the inbetween mass plays a key role. I won't lie I haven't study the double pendulum equations at all. My intuition says that it plays an important role cause it keeps the string stressed.
As post #1 shows, if we compare a simple pendulum with a mass moving horizontally on a spring, the restorative torque or force is proportional to the displacement angle in both cases. We can model the spring case with an equivalent simple pendulum. Therefore the arrangement in post #1 is akin to a double pendulum.

As post #1 shows, if we compare a simple pendulum with a mass moving horizontally on a spring, the restorative torque or force is proportional to the displacement angle in both cases. We can model the spring case with an equivalent simple pendulum. Therefore the arrangement in post #1 is akin to a double pendulum.
I don't think it is the same as a double pendulum both regarding the restorative forces and the radial forces. As to the radial forces it is obvious that it isn't the same, but also regarding the restorative forces, in the double pendulum both masses seem to swing to large angles (at least that's what i saw from the wikipedia animation) and in large angles the restorative forces becomes 2D, it isn't just horizontal as is the restorative force from that spring. And also we can't make the approximation ##sin\theta\approx \theta## for large angles, while the force from the spring implies that such an approximation is imposed.

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The equation for the torque in post #1 seems to argue that it is not a string but a rigid rod.

The equation for the torque in post #1 seems to argue that it is not a string but a rigid rod.
Yes, I think we can agree that is the intention. Posts #5 onwards have gone somewhat off topic.
The link I posted is particularly unhelpful since it does not discuss the small perturbation approximation needed for SHM, and my opinion that a string instead of a rod would be effectively the same as a double pendulum was meant to be limited to the small angle case.

Now, if it is a rigid rod, I don't see the spring remaining horizontal when the rod + bob swings through a small angle. So the extension in the spring is not just x. There is also a y component.

if it is a rigid rod, I don't see the spring remaining horizontal when the rod + bob swings through a small angle
Small is the key. The vertical displacement is of the order of θ2, so can be ignored. It's exactly the same degree of approximation we make when we claim a simple pendulum performs SHM.

• Delta2
Agreed. The change in vertical component is h ( 1 - cosθ). Is the student supposed to be aware of that?

• Delta2
Yes, I think we can agree that is the intention. Posts #5 onwards have gone somewhat off topic.
The link I posted is particularly unhelpful since it does not discuss the small perturbation approximation needed for SHM, and my opinion that a string instead of a rod would be effectively the same as a double pendulum was meant to be limited to the small angle case.

For small angles it might be the same as the double pendulum with respect to the restorative forces, but what about with respect to the radial forces?

For small angles it might be the same as the double pendulum with respect to the restorative forces, but what about with respect to the radial forces?
Likely they're different, but for small angles wouldn't it only be the restorative force that matters?