# Simple Harmonic Motion & Centripetal Force

• Kajan thana
In summary: I was wondering if there is a way to consider the restoring force as being in a plane instead of perpendicular to the string. I wasn't really sure how to go about doing that, and if it's possible at all.
Kajan thana
If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.

Kajan thana said:
If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.

Yes, but only for small ##\theta## and for circular orbits. For large ##\theta## you are no longer getting a good approximation of the motion by approximating it with motion in a plane and for non-circular orbits ##F = mv^2/r## does not apply.

Kajan thana said:
If there is a length thread with a metal ball attached at the end of the thread, and there is a oscilliation.
The restoring force is F=mgsinθ, my question is can we consider this as a centripetal force and link it to this equation: mv^2/r.

Well what kind of motion gives a centripetal force given by the constant ##mv^2/r## ?

nrqed said:
No, this force is the tangential force. The force that acts as a centripetal force is the tension in the string and this is equal to $mg cos \theta$. Note that the speed is not constant so that the value of v keeps changing.
It is the centripetal force in the horizontal plane as long as ##\theta## is small enough for this approximation to be valid. At least this is how I interpreted the question.

Edit: Also, in the planar pendulum case, the tension is not just ##mg\cos(\theta)##, but depends on the velocity as well. You will find ##T - mg \cos(\theta) = mv^2/r##.

Can you describe how you see the setup so that mg sin(θ) is centripetal force? With the appropriate approximation, of course.

Orodruin said:
It is the centripetal force in the horizontal plane as long as ##\theta## is small enough for this approximation to be valid. At least this is how I interpreted the question.

Edit: Also, in the planar pendulum case, the tension is not just ##mg\cos(\theta)##, but depends on the velocity as well. You will find ##T - mg \cos(\theta) = mv^2/r##.
Yes, you are absolutely right that the centripetal force is not mg cos theta, I said something stupid there. However, the centripetal force is not equal to mg sin theta, the centripetal force is equal to the the net force towards the center of the motion, which is T - mg cos theta.

nrqed said:
However, the centripetal force is not equal to mg sin theta, the centripetal force is equal to the the net force towards the center of the motion, which is T - mg cos theta.
Did you read my interpretation of the problem? In that interpretation ##mg \sin(\theta)## is the centripetal force! It is just with a different centre than the one you are thinking of.

nasu said:
Can you describe how you see the setup so that mg sin(θ) is centripetal force? With the appropriate approximation, of course.
I was thinking of a spherical pendulum, i.e., letting the pendulum swing in both directions and not only in a plane, as there is nothing in the OP that restricts the motion to a plane. For small angles, the curvature of the sphere that the pendulum is restricted to is well approximated by a flat surface and the restoring force within this plane is ##mg \sin(\theta)## directed towards the centre.

You can also see this by deriving the full equations of motion for the spherical pendulum and linearising them around ##\theta = 0##. You will find a linearised system equivalent to a two-dimensional centripetal problem with a centripetal force ##mg \theta = mgr/\ell##, where ##r## is the distance from the low point and ##\ell## is the length of the pendulum.

Orodruin said:
Did you read my interpretation of the problem? In that interpretation ##mg \sin(\theta)## is the centripetal force! It is just with a different centre than the one you are thinking of.I was thinking of a spherical pendulum, i.e., letting the pendulum swing in both directions and not only in a plane, as there is nothing in the OP that restricts the motion to a plane. For small angles, the curvature of the sphere that the pendulum is restricted to is well approximated by a flat surface and the restoring force within this plane is ##mg \sin(\theta)## directed towards the centre.

You can also see this by deriving the full equations of motion for the spherical pendulum and linearising them around ##\theta = 0##. You will find a linearised system equivalent to a two-dimensional centripetal problem with a centripetal force ##mg \theta = mgr/\ell##, where ##r## is the distance from the low point and ##\ell## is the length of the pendulum.
I was thinking about the usual set up for a pendulum, with the mass oscillating in a vertical plane. What made me think of my set up instead is that the OP talked about a restoring force so that, to me, this implied oscillation around the equilibrium position. I guess we have to ask him/her for more details. If you consider a general motion with theta not constant and with the oscillation not in a purely horizontal plane, it does not seem to me that we can talk about a centripetal force since the motion is not circular.

nrqed said:
If you consider a general motion with theta not constant and with the oscillation not in a purely horizontal plane, it does not seem to me that we can talk about a centripetal force since the motion is not circular.

As long as you stay at small angles, the curvature of the sphere is negligible. For small angles the problem is exactly equivalent to a two-dimensional harmonic oscillator.

Last edited:

## 1. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which an object oscillates back and forth around an equilibrium point, with a force that is directly proportional to its displacement from the equilibrium point. This motion is characterized by a sinusoidal pattern.

## 2. What is the equation for simple harmonic motion?

The equation for simple harmonic motion is F = -kx, where F represents the force, k is the spring constant, and x is the displacement from equilibrium.

## 3. What is the relationship between simple harmonic motion and centripetal force?

Simple harmonic motion and centripetal force are related in that they both involve a restoring force that acts towards an equilibrium point. However, centripetal force specifically refers to the force that keeps an object moving in a circular path, while simple harmonic motion can occur in various types of periodic motion.

## 4. How do you calculate the period of a simple harmonic motion?

The period of a simple harmonic motion can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant.

## 5. What are some real-life examples of simple harmonic motion?

Some real-life examples of simple harmonic motion include a mass attached to a spring, a pendulum, a swinging door, and a guitar string. All of these examples involve a restoring force that causes the object to oscillate back and forth around an equilibrium point.

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