# Oscillating simple pendulumWatch

#1
'Explain the physics of why the tension in the string of an oscillating simple pendulum is not equal to the weight of the pendulum bob when the string is vertical.'

I was thinking that as the pendulum is oscillating, say it had swung from the right hand side to reach the middle, the string is vertical but the bob is accelerating to the left hand side, so the resultant force must be acting in that direction. This means the tension in the string wouldn't equal the weight of the bob. Am I correct in thinking that? If so, why is there a resultant force acting to the left?
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5 years ago
#2
(Original post by dr-jimmy)
'Explain the physics of why the tension in the string of an oscillating simple pendulum is not equal to the weight of the pendulum bob when the string is vertical.'

I was thinking that as the pendulum is oscillating, say it had swung from the right hand side to reach the middle, the string is vertical but the bob is accelerating to the left hand side, so the resultant force must be acting in that direction. This means the tension in the string wouldn't equal the weight of the bob. Am I correct in thinking that? If so, why is there a resultant force acting to the left?
The bits in bold are not correct. If you drop a pendulum from the right it will accelerate to the left only until it reaches the middle (equilibrium position) at which point it will accelerate to the right, decreasing its velocity to the left until the point where it is at maximum displacement to the left and 0 velocity and 0 acceleration.

Remember that acceleration is always to the equilibrium position and therefore displacement is always in the other direction to acceleration. When displacement is 0, acceleration is 0.
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#3
(Original post by YThursday)
The bits in bold are not correct. If you drop a pendulum from the right it will accelerate to the left only until it reaches the middle (equilibrium position) at which point it will accelerate to the right, decreasing its velocity to the left until the point where it is at maximum displacement to the left and 0 velocity and 0 acceleration.

Remember that acceleration is always to the equilibrium position and therefore displacement is always in the other direction to acceleration. When displacement is 0, acceleration is 0.
Okay, that makes sense, thanks.
So why does the tension in the string not equal the weight of the bob when it reaches the equilibrium position?
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5 years ago
#4
(Original post by dr-jimmy)
'Explain the physics of why the tension in the string of an oscillating simple pendulum is not equal to the weight of the pendulum bob when the string is vertical.'

I was thinking that as the pendulum is oscillating, say it had swung from the right hand side to reach the middle, the string is vertical but the bob is accelerating to the left hand side, so the resultant force must be acting in that direction. This means the tension in the string wouldn't equal the weight of the bob. Am I correct in thinking that? If so, why is there a resultant force acting to the left?
very simply the bob doesn't just oscillate in the left and right axis, it's also going up and down. At the vertical position it's accelerating upwards and the force to cause that acceleration must come from the tension in the string.
You might want to think again about what the horizontal component of the acceleration is at this point as I don't think you have it right in your explanation above.
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#5
(Original post by Joinedup)
very simply the bob doesn't just oscillate in the left and right axis, it's also going up and down. At the vertical position it's accelerating upwards and the force to cause that acceleration must come from the tension in the string.
You might want to think again about what the horizontal component of the acceleration is at this point as I don't think you have it right in your explanation above.
So when the bob reaches the equilibrium position, the resultant force is vertically upwards? So the tension in the string is greater than the weight of the bob? Why is that?
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5 years ago
#6
The bob travels in an arc, i.e. there is circular motion.

Hence, for the bob not to continue moving at a tangent to its path, there must be acceleration towards the center of the setup.

Think of just the vertical components. If the weight = tension, then the bob would stay at a constant height, as there is no resultant force up or down.

Since the bob does move up, however, there is acceleration upwards and therefore the tension is greater than the weight.
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5 years ago
#7
(Original post by dr-jimmy)
So when the bob reaches the equilibrium position, the resultant force is vertically upwards? So the tension in the string is greater than the weight of the bob? Why is that?
the instant in time before reaching equilibrium position the bob was moving downwards and the instant after passing the equilibrium position the bob is moving upwards so it has undergone acceleration.
F=ma
it's a centrepetal force that's acting vertically at that instant.
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5 years ago
#8
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#9
(Original post by Joinedup)
the instant in time before reaching equilibrium position the bob was moving downwards and the instant after passing the equilibrium position the bob is moving upwards so it has undergone acceleration.
F=ma
it's a centrepetal force that's acting vertically at that instant.
(Original post by Albino_muffin)
The bob travels in an arc, i.e. there is circular motion.

Hence, for the bob not to continue moving at a tangent to its path, there must be acceleration towards the center of the setup.

Think of just the vertical components. If the weight = tension, then the bob would stay at a constant height, as there is no resultant force up or down.

Since the bob does move up, however, there is acceleration upwards and therefore the tension is greater than the weight.
So there is a centripetal force, which always acts toward the centre of the circle? Is centripetal force always a resultant force? Which forces have been resolved to give the resultant force that causes the bob to accelerate upwards at the equilibrium position?
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5 years ago
#10
(Original post by dr-jimmy)
So there is a centripetal force, which always acts toward the centre of the circle? Is centripetal force always a resultant force? Which forces have been resolved to give the resultant force that causes the bob to accelerate upwards at the equilibrium position?
There are 2 forces on the bob, its weight (W downwards) and the tension (T upwards) in the string.
The resultant of these two must equal the centripetal force F.

Write the equation for this and rearrange for T.

What does it tell you?
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#11
(Original post by Stonebridge)
There are 2 forces on the bob, its weight (W downwards) and the tension (T upwards) in the string.
The resultant of these two must equal the centripetal force F.

Write the equation for this and rearrange for T.

What does it tell you?
WT = F
T = F/W

I'm not sure what it tells me. The only way the tension in the string can be greater than the weight of the bob is for F to be greater than W^2? Sorry I don't know what to infer from the equation.
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5 years ago
#12
(Original post by dr-jimmy)
WT = F
T = F/W

I'm not sure what it tells me. The only way the tension in the string can be greater than the weight of the bob is for F to be greater than W^2? Sorry I don't know what to infer from the equation.
Where do you get WT=F from?

The equation is
T - W = F
T is up (positive)
W is down (negative)
F, the centripetal force, is upwards towards to the centre (positive)
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#13
(Original post by Stonebridge)
Where do you get WT=F from?

The equation is
T - W = F
T is up (positive)
W is down (negative)
F, the centripetal force, is upwards towards to the centre (positive)
LOL fml I don't even know. I think I was thinking of when you have like the dot product of two vectors so multiplying them? Still stupid I won't bother trying to justify it.
Why does the centripetal force always face the centre?
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5 years ago
#14
(Original post by dr-jimmy)
LOL fml I don't even know. I think I was thinking of when you have like the dot product of two vectors so multiplying them? Still stupid I won't bother trying to justify it.
Why does the centripetal force always face the centre?
Because it has to be in the same direction as the acceleration. How can a (resultant) force do anything other than produce an acceleration in the same direction?
The acceleration is always towards the centre of the circle for an object moving in a circle.
Motion in a circle is caused by a constant force acting towards the centre of the circle.
At the point where the pendulum is at the bottom, you have the case where the forces all act in the same line. Hence the nice easy formula in the previous post, which shows why the tension must be greater than the weight at that point.
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#15
(Original post by Stonebridge)
Because it has to be in the same direction as the acceleration. How can a (resultant) force do anything other than produce an acceleration in the same direction?
The acceleration is always towards the centre of the circle for an object moving in a circle.
Motion in a circle is caused by a constant force acting towards the centre of the circle.
At the point where the pendulum is at the bottom, you have the case where the forces all act in the same line. Hence the nice easy formula in the previous post, which shows why the tension must be greater than the weight at that point.
I get that it has to be in the same direction as the acceleration. I think what I meant to ask is why the acceleration is always towards the centre of the circle. Thanks for your patience btw.
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5 years ago
#16
(Original post by dr-jimmy)
I get that it has to be in the same direction as the acceleration. I think what I meant to ask is why the acceleration is always towards the centre of the circle. Thanks for your patience btw.
If the object is moving at constant speed in a circle, then the acceleration only changes it's direction. The acceleration (and force) must always be at right angles to the motion, because if it wasn't it would have a component that caused the object to speed up or slow down. It's only an acceleration at right angles to the motion that can keep it moving at a constant speed yet changing direction. By definition, the line at right angles to the motion points to the centre of the circle.
If you want a more mathematical proof do a google search for it. There are plenty out there. Try Youtube.
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#17
(Original post by Stonebridge)
If the object is moving at constant speed in a circle, then the acceleration only changes it's direction. The acceleration (and force) must always be at right angles to the motion, because if it wasn't it would have a component that caused the object to speed up or slow down. It's only an acceleration at right angles to the motion that can keep it moving at a constant speed yet changing direction. By definition, the line at right angles to the motion points to the centre of the circle.
If you want a more mathematical proof do a google search for it. There are plenty out there. Try Youtube.
Okay thanks. I looked up a proof and it makes sense.
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