Which one is a correct statement about oscillation?

[songoku]

Homework Statement

Please see below

Relevant Equations

Simple Harmonic Motion
$a=-\omega^2 x$
$F=m.a$

The answer is (C), but in my opinion all options are wrong:
(A) is wrong because at X and Y the net force is maximum, not zero
(B) is wrong because the direction of net force can be in opposite direction to velocity
(C) is wrong because at the middle (equilibrium point) the acceleration is zero
(D) is wrong because they are not directly proportional

Why is (C) correct?

Thanks

 

[TSny]

(C) is wrong because at the middle (equilibrium point) the acceleration is zero

Be careful. Can a particle have nonzero acceleration at an instant when its speed is not changing?

 

[TSny]

A pendulum will approximate simple harmonic motion only for small oscillations. The problem statement does not indicate small oscillations. So, you cannot assume the motion is simple harmonic. Nevertheless, you should be able to argue that only (C) is correct.

 

[TSny]

(A) is wrong because at X and Y the net force is maximum, not zero

You are correct that the net force is not zero at X or Y. (However, for large enough oscillations the net force will not be maximum at these two points.)

Can you explain why the net force can't be zero at X or Y?

 

[haruspex]

(B) is wrong because the direction of net force can be in opposite direction to velocity

As @TSny points out, you should not be assuming SHM here.

(D) is wrong because they are not directly proportional

That’s like saying it's wrong because it’s wrong.

 

[TSny]

(B) is wrong because the direction of net force can be in opposite direction to velocity

The net force on the bob will never have a direction opposite to the velocity direction.

Try to see why the net acceleration and velocity never have opposite directions for the pendulum. (However, for small oscillations, they can be approximately opposite.)

 

[kuruman]

(C) is wrong because at the middle (equilibrium point) the acceleration is zero

A pendulum bob moves continuously along a circular arc between two turning points. Can you provide a simple argument why an object in continuous motion along a curved path must necessarily have non zero acceleration? The argument works for any curved path, not only a circular arc.

This animation from the pendulum Wikipedia article is quite informative. Note the relation between the acceleration and the velocity directions. You can also see that the velocity goes through zero at the turning points. However, the acceleration is never zero.

Something is zero at the lowest point of the motion, but it is not the net force. What quantity (other than the angular displacement $\theta$) is it?

Edited for typo, see post #9.

 

[songoku]

A pendulum will approximate simple harmonic motion only for small oscillations. The problem statement does not indicate small oscillations. So, you cannot assume the motion is simple harmonic. Nevertheless, you should be able to argue that only (C) is correct.

As @TSny points out, you should not be assuming SHM here.

Be careful. Can a particle have nonzero acceleration at an instant when its speed is not changing?

Ah ok, so that's my mistake.

Since this is not SHM, so the motion of the bob can't be taken as straight horizontal line but will be an arc so I think there will be two accelerations; one is from the restoring force and one is centripetal acceleration

At the middle, the acceleration from restoring force is zero but there is centripetal acceleration so even though the speed is not changing, the acceleration is non zero.

That’s like saying it's wrong because it’s wrong.

It is because when velocity is increasing, acceleration is decreasing so they are not directly proportional.

You are correct that the net force is not zero at X or Y. (However, for large enough oscillations the net force will not be maximum at these two points.)

Can you explain why the net force can't be zero at X or Y?

The net force can't be zero at X or Y because there will be restoring force pulling the pendulum back to opposite direction but I don't know why the force is not maximum at X or Y

The net force on the bob will never have a direction opposite to the velocity direction.

Try to see why the net acceleration and velocity never have opposite directions for the pendulum. (However, for small oscillations, they can be approximately opposite.)

I think the answer is in the diagram posted by @kuruman

A pendulum bob moves continuously along a circular arc between two turning points. Can you provide a simple argument why an object in continuous motion along a curved path must necessarily have zero acceleration? The argument works for any curved path, not only a circular arc.

Maybe you mean "must necessarily have non-zero acceleration"?

I suppose the reason is because the direction of motion is changing so velocity is changing. Rate of change of velocity is acceleration

Something is zero at the lowest point of the motion, but it is not the net force. What quantity (other than the angular displacement $\theta$) is it?

Gravitational potential energy?

 

[kuruman]

Maybe you mean "must necessarily have non-zero acceleration"?

Yes. that's what I meant. I fixed the typo. Thanks for the catch and sorry about the confusion.

Gravitational potential energy?

The zero of gravitational potential is arbitrary and can be chosen anywhere. What two vector quantities must be zero for an object to be in equilibrium? One is the net force. What is the other one?

 

[Mister T]

The author of the question was likely thinking (correctly) that the tangential component of the acceleration is zero at the equilibrium point and wasn't thinking about the nonzero centripetal component. It's a common error.

 

[jbriggs444]

It is because when velocity is increasing, acceleration is decreasing so they are not directly proportional.

As written, this is not exactly correct.

Linear damping (e.g. simple viscous drag) is a case where acceleration becomes increasingly negative as velocity becomes increasingly positive.

That said, an observation that the magnitude of velocity is increasing while the magnitude of acceleration is decreasing does indicate a lack of direct proportionality.

 

[TSny]

Since this is not SHM, so the motion of the bob can't be taken as straight horizontal line but will be an arc so I think there will be two accelerations; one is from the restoring force and one is centripetal acceleration

At the middle, the acceleration from restoring force is zero but there is centripetal acceleration so even though the speed is not changing, the acceleration is non zero.

Yes.

The net force can't be zero at X or Y because there will be restoring force pulling the pendulum back to opposite direction

Yes.

but I don't know why the force is not maximum at X or Y

Take an extreme case where the bob is released with the string horizontal. Use energy conservation to find the speed at the bottom of the swing. Compare the centripetal acceleration at the bottom to the acceleration at X or Y. Compare the net force on the bob at X to the net force at the bottom.

 

[songoku]

The zero of gravitational potential is arbitrary and can be chosen anywhere. What two vector quantities must be zero for an object to be in equilibrium? One is the net force. What is the other one?

Should be net torque

Take an extreme case where the bob is released with the string horizontal. Use energy conservation to find the speed at the bottom of the swing. Compare the centripetal acceleration at the bottom to the acceleration at X or Y. Compare the net force on the bob at X to the net force at the bottom.

speed at bottom of the swing = $\sqrt{2gL}$ where $L$ is length of pendulum

Net force at bottom = $T-W$ where $T$ is the tension and $W$ is the weight of pendulum bob
Centripetal acceleration at bottom = $\frac{T-W}{m}$

acceleration at X or Y is due to restoring force. Is restoring force the resultant of tension and weight at this position?

 

[kuruman]

Should be net torque

Yes.

Is restoring force the resultant of tension and weight at this position?

The net force is the resultant of the tension and the weight. It generates the restoring torque (not force) about the point of support.

 

[TSny]

speed at bottom of the swing = $\sqrt{2gL}$ where $L$ is length of pendulum

OK. So, what is the centripetal acceleration at the bottom expressed in terms of g?
What is the net force on the bob at the bottom expressed in terms of m and g?

When the bob is at X (where we are taking the string to be horizontal in this case), what is the acceleration of the bob? What is the net force on the bob at X expressed in terms of m and g?

Compare the net force at X with the net force at the bottom. You should see that the net force is not at a maximum at X.

 

[songoku]

The net force is the resultant of the tension and the weight. It generates the restoring torque (not force) about the point of support.

Why can't I say it generates restoring force?

OK. So, what is the centripetal acceleration at the bottom expressed in terms of g?

2g

What is the net force on the bob at the bottom expressed in terms of m and g?

2mg

When the bob is at X (where we are taking the string to be horizontal in this case), what is the acceleration of the bob?

I suppose it would be g

What is the net force on the bob at X expressed in terms of m and g?

mg

Compare the net force at X with the net force at the bottom. You should see that the net force is not at a maximum at X.

Yes, the net force actually is bigger at the bottom. Is bottom the position where the force is maximum?

 

[kuruman]

Why can't I say it generates restoring force?

Because a torque is required to restore the angular displacement to equilibrium.

 

[songoku]

Because a torque is required to restore the angular displacement to equilibrium.

Is the term restoring force only appropriate to be used for SHM case?

 

[kuruman]

Is the term restoring force only appropriate to be used for SHM case?

Not necessarily. The term "harmonic" simply means that sines and cosines are used to describe the motion. It is the case when the restoring force (or torque) is proportional to the negative of the linear (or angular) acceleration. Read about anharmonicity here.

 

[TSny]

2g

2mg

I suppose it would be g

mg

Yes, the net force actually is bigger at the bottom.

Yes to all of the above.

Is bottom the position where the force is maximum?

You can show that if the angular amplitude of the swing is greater than $\tan^{-1}(4/3) \approx 53^\circ$, the maximum force during the oscillation occurs at the bottom of the swing.

For amplitudes less than this, max force occurs at X (or Y).

 

[TSny]

Here's an interactive animation. Click on "Lab" to explore the acceleration for different amplitudes.

 

[songoku]

Thank you very much for all the help and explanation TSny, kuruman, haruspex, Mister T, jbriggs444

 

[jbriggs444]

Why can't I say it generates restoring force?

If you look at the problem in one dimension (tangential) by discarding the irrelevant constraint force in the radial direction then yes, one could regard the tangential component of gravity as a "restoring force".

Whether radial force or torque from the pendulum arm are truly "irrelevant" may depend on the exact quantity one is trying to determine and the level of detail of the model.

 

[kuruman]

If you look at the problem in one dimension (tangential) by discarding the irrelevant constraint force in the radial direction then yes, one could regard the tangential component of gravity as a "restoring force".

Proof
If a restoring force is to be identified, then one should look for it as the negative gradient of the potential energy in which the mass finds itself. Here we have $$U=mgR(1-\cos\!\theta)$$. The potential energy is a function of $\theta$ only in which case $$\mathbf F=-\frac{1}{R}\frac{\partial U}{\partial \theta}\mathbf{\hat {\theta}}=-mg\sin\!\theta~\mathbf{\hat {\theta}}.$$

 

[Mister T]

Is the term restoring force only appropriate to be used for SHM case?

No, it's just that SHM is characterized by a linear restoring force.