Frequency of pendulum in the presence of electric field

[songoku]

Homework Statement

A tiny neutral sphere with mass 𝑚 is suspended from a massless string of length 𝐿 in the presence of a uniform electric field 𝐸 pointing in +y direction. The ball undergoes small oscillations, exhibiting simple harmonic motion with a known frequency of $\omega=\sqrt{\frac{g}{L}}$. Upon introducing a positive charge 𝑞 to the sphere, what will happen to the frequency? It is assumed that 𝑚𝑔 > 𝑞𝐸.
A) increase
B) decrease
C) remain the same
D) the motion is no longer simple harmonic

Relevant Equations

F = Eq

F = ma

$a=-\omega^2 x$

I imagine the setup to be like this: the pendulum is moving horizontally, then it will experience electric force upwards so there is additional upwards acceleration, in addition to the acceleration from restoring force.

Since the resultant of the two accelerations is not directed horizontally towards the equilibrium point, I thought the acceleration is not directly proportional to displacement anymore so the answer would be (D), but the correct answer is (B)

What is my mistake?

Thanks

 

[haruspex]

Homework Statement: A tiny neutral sphere with mass 𝑚 is suspended from a massless string of length 𝐿 in the presence of a uniform electric field 𝐸 pointing in +y direction. The ball undergoes small oscillations, exhibiting simple harmonic motion with a known frequency of $\omega=\sqrt{\frac{g}{L}}$. Upon introducing a positive charge 𝑞 to the sphere, what will happen to the frequency? It is assumed that 𝑚𝑔 > 𝑞𝐸.
A) increase
B) decrease
C) remain the same
D) the motion is no longer simple harmonic
Relevant Equations: F = Eq

F = ma

$a=-\omega^2 x$

I imagine the setup to be like this: the pendulum is moving horizontally, then it will experience electric force upwards so there is additional upwards acceleration, in addition to the acceleration from restoring force.

Since the resultant of the two accelerations is not directed horizontally towards the equilibrium point, I thought the acceleration is not directly proportional to displacement anymore so the answer would be (D), but the correct answer is (B)

What is my mistake?

Thanks

I can see argument for (B) but I am unconvinced. I feel we need to be careful, and not assume it can be treated as purely horizontal motion.
With that in mind, which way does the electric force act?

 

[songoku]

I can see argument for (B) but I am unconvinced. I feel we need to be careful, and not assume it can be treated as purely horizontal motion.

Although the question has already stated that the motion is SHM, we still can't assume it to be purely horizontal motion?

With that in mind, which way does the electric force act?

upward

 

[haruspex]

Although the question has already stated that the motion is SHM, we still can't assume it to be purely horizontal motion?

It clearly cannot be purely horizontal, no matter how small the perturbation. Much safer to treat it accurately at first, only making approximations later.

upward

What causes the force? Quote an equation.

 

[songoku]

What causes the force? Quote an equation.

The electric field causes the electric force. The equation is F = Eq

 

[haruspex]

The electric field causes the electric force. The equation is F = Eq

I'm sorry… I misread it as magnetic field and took the y direction as into the page, z being vertical. Let me start again…

Right, the electric force is purely vertical. What other force acting is vertical?

 

[Orodruin]

If you treat it as purely horizontal motion then there is no oscillation even in the pure gravity case …

In this case, the gravitational force in the y direction is replaced by a gravitational and electric force $-mg \to -mg + qE = -m(g - qE/m)$. The implications for the frequency should be straightforward.

 

[songoku]

Right, the electric force is purely vertical. What other force acting is vertical?

Weight and vertical component of tension

If you treat it as purely horizontal motion then there is no oscillation even in the pure gravity case …

In this case, the gravitational force in the y direction is replaced by a gravitational and electric force $-mg \to -mg + qE = -m(g - qE/m)$. The implications for the frequency should be straightforward.

We don't need to consider tension?

 

[Orodruin]

We don't need to consider tension?

Tension is what it needs to be to maintain circular motion. Just as in the case with only a gravitational field.

 

[haruspex]

Weight and vertical component of tension

The easiest way to look at this, as @Orodruin describes, is to note that adding a constant upward force is equivalent to reducing g.

Another way is to resolve into forces along the pendulum arm and normal to it. The tension acts along the arm of the pendulum, which means it has no component in the direction of the velocity of the mass. Thus, it does not feature in the equation for the angular acceleration, so does not affect the period. Likewise, the components of g and the electric force that act in that direction. That leaves their components normal to the pendulum, and these oppose, so the electric force reduces the restoring force.

 

[songoku]

Tension is what it needs to be to maintain circular motion. Just as in the case with only a gravitational field.

The easiest way to look at this, as @Orodruin describes, is to note that adding a constant upward force is equivalent to reducing g.

Another way is to resolve into forces along the pendulum arm and normal to it. The tension acts along the arm of the pendulum, which means it has no component in the direction of the velocity of the mass. Thus, it does not feature in the equation for the angular acceleration, so does not affect the period. Likewise, the components of g and the electric force that act in that direction. That leaves their components normal to the pendulum, and these oppose, so the electric force reduces the restoring force.

I think the condition mg > qE is so that the resultant force still directed to equilibrium point so it can act as restoring force

Is it correct to say that if mg < qE then the motion is not SHM anymore?

Thanks

 

[Orodruin]

I think the condition mg > qE is so that the resultant force still directed to equilibrium point so it can act as restoring force

Is it correct to say that if mg < qE then the motion is not SHM anymore?

Thanks

It is never SHM, just approximately for small oscillations around the equilibrium.

If mg < qE the pendulum equilibrium would instead be in the up direction. Small oscillations around that new equilibrium would also be approximately SHM.

 

[songoku]

If mg < qE the pendulum equilibrium would instead be in the up direction. Small oscillations around that new equilibrium would also be approximately SHM.

Is it possible for the pendulum to oscillate vertically in this case? I think the tension will need to be considered in this case so the tension will affects the motion to make it not SHM?

Thanks

 

[haruspex]

Is it possible for the pendulum to oscillate vertically in this case? I think the tension will need to be considered in this case so the tension will affects the motion to make it not SHM?

Thanks

@Orodruin means the entire pendulum is inverted. The charged mass is oscillating in a small arc distance L above the point of attachment.

 

[songoku]

@Orodruin means the entire pendulum is inverted. The charged mass is oscillating in a small arc distance L above the point of attachment.

Oh ok. But let say in initial scenario the pendulum is attached to rigid and really large ceiling so upon introducing a positive charge 𝑞 to the sphere and mg < qE, is it correct that the sphere will not oscillate but just rotate until it hits the ceiling?

Thanks

 

[haruspex]

Oh ok. But let say in initial scenario the pendulum is attached to rigid and really large ceiling so upon introducing a positive charge 𝑞 to the sphere and mg < qE, is it correct that the sphere will not oscillate but just rotate until it hits the ceiling?

Thanks

Yes.

 

[songoku]

Thank you very much for the help and explanation haruspex and Orodruin