How Does a Spring Impact the Angular Frequency of a Pendulum?

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.
  • #1
Ujjwal Basumatary
17
0

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.
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  • #2
Hello. Your work looks correct to me.
 
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  • #3
Is the bob suspended by a flexible string or a rigid rod?
 
  • #4
Ujjwal Basumatary said:
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.
 
  • #5
Chandra Prayaga said:
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?
 
  • #7
haruspex said:
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.
 
  • #8
Delta² said:
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.
 
  • #9
haruspex said:
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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  • #10
The equation for the torque in post #1 seems to argue that it is not a string but a rigid rod.
 
  • #11
Chandra Prayaga said:
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.
 
  • #12
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.
 
  • #13
Chandra Prayaga said:
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.
 
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  • #14
Agreed. The change in vertical component is h ( 1 - cosθ). Is the student supposed to be aware of that?
 
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  • #15
haruspex said:
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?
 
  • #16
Delta² said:
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?
 

Related to How Does a Spring Impact the Angular Frequency of a Pendulum?

1. What is a spring attached to a pendulum?

A spring attached to a pendulum is a physical system where a spring is attached to the end of a pendulum. This creates a more complex pendulum motion due to the added force and energy from the spring.

2. How does a spring affect the motion of a pendulum?

A spring attached to a pendulum adds an additional force to the motion, causing the pendulum to oscillate with a combination of both the gravitational force and the elasticity of the spring. This creates a more complex and varied motion compared to a simple pendulum.

3. What are the factors that affect the motion of a spring attached to a pendulum?

The factors that affect the motion of a spring attached to a pendulum include the length of the pendulum, the mass of the pendulum, the strength and stiffness of the spring, and the initial displacement or angle at which the pendulum is released.

4. How does the spring constant affect the motion of a spring attached to a pendulum?

The spring constant, also known as the stiffness of the spring, determines how much force is needed to stretch or compress the spring. A higher spring constant will result in a stronger force and a more energetic motion of the pendulum, while a lower spring constant will result in a weaker force and a less energetic motion.

5. What is the difference between a simple pendulum and a spring attached to a pendulum?

The main difference between a simple pendulum and a spring attached to a pendulum is the added force and energy from the spring. A simple pendulum follows a path based on the gravitational force alone, while a spring attached to a pendulum has a more complex motion due to the combined forces of gravity and the spring's elasticity.

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