Calculating Time Period of Simple Pendulum with Charged Capacitor | SHM Homework

[Amith2006]

Homework Statement

# A simple pendulum having charge q, mass m and effective length l is suspended from a rigid support between the plates of a charged capacitor whose plates are kept vertical. What is the time period of oscillation of the pendulum?

Homework Equations

Time period = 2(pi)[l/g’]^(1/2)

The Attempt at a Solution

I solved it in the following way:
Let x be the angular displacement at any instant. Let T be the tension in the string.
Restoring force(F) = -(mgsinx + qEcosx)
For small values of x,
F= -(mgx + qE)
But this won’t be a Simple harmonic motion. I am stuck here.
The solution as given in my book is as follows:
Tcosx = mg ---------(1)
Tsinx = qE ---------(2)
From (1) & (2),
T = [(mg)^2 + (qE)^2]^(1/2)
Effective g’= T/m
= [g^2 + (qE/m)^2]^(1/2)
Time period = 2(pi)[l/g’]^(1/2)
= 2(pi)[l/(g^2 + (qE/m)^2)^(1/2)]^(1/2)

As per them, the direction of effective g’ is inclined to the vertical by angle x. But effective g’ should be vertical, isn’t it? Does tension contribute to the restoring force experienced by the bob?I think it is the component of weight that contributes to the restoring force? But they haven’t used that concept at all to find the time period. Please help!

 

[andrevdh]

The weight of an object can be determined by hanging it from a spring scale. In such a case we say that the weight of the object is equal to the tension in the spring. The "effective g" would then be the tension in the spring divided by the mass hanging from it. (The derivation of the period of a simple pendulum is done under the assumption that the pendulum swings only slightly out of the bottom position, which means that the speed of the bob will be small resulting in a negligble centripetal acceleration - the tension therefore remains virtually the same).

 

[chaoseverlasting]

Forces acting are $$F_e=qE$$ and $$F_g=mg$$. Find the net acceleration due to the two forces and that will be your effective g. Then use that term instead of g in your expression for the time period.

 

[chaoseverlasting]

Homework Statement

Restoring force(F) = -(mgsinx + qEcosx)
For small values of x,
F= -(mgx + qE)
But this won’t be a Simple harmonic motion. I am stuck here.

To find the DE for SHM, find the torque about the point of suspension due to mg (torque due to tension will be zero as its line of action passes through the point of suspension) and equate that to $$I\alpha$$ where $$\alpha=\frac{d^2x}{dt^2}$$. Here you can use the assumption sinx=x (for small values) and you'll get your DE.

As per them, the direction of effective g’ is inclined to the vertical by angle x. But effective g’ should be vertical, isn’t it?

Here effective g really means the net acceleration acting on the bob. Dont think its the acceleration in the downward direction.

Does tension contribute to the restoring force experienced by the bob?I think it is the component of weight that contributes to the restoring force? But they haven’t used that concept at all to find the time period. Please help!

Yes, tension does contribute to the restoring force. Infact, tension here is the restoring force. That is why the expression for g(eff) is T/m.
F(restoring)=ma
T=F(restoring)=mg(eff)
Therefore, g(eff)=F(restoring)/m=T/m

 

[Amith2006]

To find the DE for SHM, find the torque about the point of suspension due to mg (torque due to tension will be zero as its line of action passes through the point of suspension) and equate that to $$I\alpha$$ where $$\alpha=\frac{d^2x}{dt^2}$$. Here you can use the assumption sinx=x (for small values) and you'll get your DE.

Yes, tension does contribute to the restoring force. Infact, tension here is the restoring force. That is why the expression for g(eff) is T/m.

You said that torque due to tension is zero. In that case will tension contribute to the restoring force?Isn't what you say contradictory?

 

[andrevdh]

Surely $$F_E$$ also has a torque contribution and should be included in the derivation of the DE!

To elaborate further on my previous point - the effective weight of an object can be defined as the force it exerts on a spring scale in a particular reference system. Therefore the effective gravitational acceleration follows from

$$g\prime = \frac{T}{m}$$

where $$T$$ is the reading on the spring scale. The direction of the effective gravitational acceleration of the object is then in the opposite direction of the tension vector.

 

[Amith2006]

That's cool!Thanks.

 

[chaoseverlasting]

Yeah, sorry. $$F_e$$ must also be included in the expression for torque.

 

[chaoseverlasting]

Torque of the tension about the point of suspension is zero, but that doesn't mean that it doesn't contribute to the restoring force.

Torque can be zero, but the force necessarily isnt. Similarly, the net force can be zero, but the torque doesn't have to be (electric dipole in a uniform electric field).

The torque is zero, because the line of action of the force passes through that point. Since torque is defined as (force)x(perpendicular distance), the perpendicular distance is zero, hence the torque is zero.