Accelerated Pendulum Frequency: g-a vs g+a

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In summary, the frequency of a pendulum is proportional to the square root of the acceleration due to gravity divided by the length of the pendulum. When the pendulum is accelerated upward by a force, the effective downward acceleration is (g+a), taking into account the pseudo force caused by the acceleration of the peg holding the pendulum.
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Assume a pendulum, a mass M attached to a massless rod, which is oscillating in small angle. According to the oscillating equation

[tex]\ddot{\theta} + \frac{g}{l}\theta = 0[/tex]

We find that the frequency is proportional to [tex]\sqrt{g/l}[/tex]. Now if the pendulum is accelerated upward at acceleration [tex]a[/tex], since [tex]a[/tex] is along the opposite direction as the gravity [tex]g[/tex], the frequency of the accelerated pendulum should be

[tex]\omega = \sqrt{\frac{g-a}{l}}[/tex]

why in textbook, in said the frequency becomes [tex]\omega = \sqrt{\frac{g+a}{l}}[/tex]?
 
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Aha! Actually, it is not the pendulum that is accelerated upwards! It is the peg that is holding the pendulum that is accelerated. So the pendulum itself experiences a pseudo force downwards.

If you stand in an elevator and it accelerates at "a" then your weight
as you would experience will be m(g+a), because ma downwards is the pseudo force.

Pendulum is no different: the effective downward acceleration of the pendulum of mass m is (mg + pseudo force)/m = (mg + ma)/m = (g+a).

Makes sense?
 
  • #3
sai_2008 said:
Aha! Actually, it is not the pendulum that is accelerated upwards! It is the peg that is holding the pendulum that is accelerated. So the pendulum itself experiences a pseudo force downwards.

If you stand in an elevator and it accelerates at "a" then your weight
as you would experience will be m(g+a), because ma downwards is the pseudo force.

Pendulum is no different: the effective downward acceleration of the pendulum of mass m is (mg + pseudo force)/m = (mg + ma)/m = (g+a).

Makes sense?

Yes, it makes sense.
 

FAQ: Accelerated Pendulum Frequency: g-a vs g+a

What is an accelerated pendulum?

An accelerated pendulum is a device consisting of a weight, or bob, attached to a string or rod that is able to swing freely back and forth due to the force of gravity.

What is the frequency of an accelerated pendulum?

The frequency of an accelerated pendulum refers to the number of complete swings, or oscillations, it makes per unit of time. This is typically measured in hertz (Hz).

How is the frequency of an accelerated pendulum affected by the acceleration of gravity?

The frequency of an accelerated pendulum is directly affected by the acceleration of gravity. As the acceleration of gravity increases, the frequency of the pendulum also increases.

What is the difference between g-a and g+a in relation to the frequency of an accelerated pendulum?

g-a and g+a refer to the direction of acceleration in relation to gravity. g-a means that the acceleration is in the opposite direction of gravity, while g+a means that the acceleration is in the same direction as gravity. This difference can affect the frequency of the pendulum.

How is the frequency of an accelerated pendulum measured?

The frequency of an accelerated pendulum can be measured by counting the number of oscillations it makes in a given amount of time, or by using a stopwatch to measure the time it takes for a certain number of oscillations to occur.

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