Static extension - Simple Harmonic Motion
Hi, could someone explain it to me what static extension (denote it for letter e) is and how does it differ to x when k = mg/x to work out the spring constant?
Also, as for the second part, the way I would have approached is simply used the equation Maximum Potential Energy = 0.5k(A^2) where potential energy is at its maximum at the amplitude (A). This approach is incorrect and requires a bit of extra working out.
The answer is meant to be (0.5)(k)((e+A)^2) = 0.56J
So why do we add e to the A?
Thank you very much.
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You first put the weight on the spring and it extends e.
So use this for the spring constant as you have the tension in the spring (weight of mass) and its extension.
You then pull the weight down an extra amount x and release it.
Total extension now e+x
x is then the amplitude of the oscillations.
At maximum displacement for the SHM the spring is stretched e+x downwards and e-x upwards.
Max eleastic PE in spring occurs at max extension downwards.
So use this for the spring constant as you have the tension in the spring (weight of mass) and its extension.
You then pull the weight down an extra amount x and release it.
Total extension now e+x
x is then the amplitude of the oscillations.
At maximum displacement for the SHM the spring is stretched e+x downwards and e-x upwards.
Max eleastic PE in spring occurs at max extension downwards.
(edited 10 years ago)
Original post by Stonebridge
You first put the weight on the spring and it extends e.
So use this for the spring constant as you have the tension in the spring (weight of mass) and its extension.
You then pull the weight down an extra amount x and release it.
Total extension now e+x
x is then the amplitude of the oscillations.
At maximum displacement for the SHM the spring is stretched e+x downwards and e-x upwards.
Max eleastic PE in spring occurs at max extension downwards.
So use this for the spring constant as you have the tension in the spring (weight of mass) and its extension.
You then pull the weight down an extra amount x and release it.
Total extension now e+x
x is then the amplitude of the oscillations.
At maximum displacement for the SHM the spring is stretched e+x downwards and e-x upwards.
Max eleastic PE in spring occurs at max extension downwards.
Hi, thank you very much for your response and effort. I greatly appreciate it.
I have sketched an image according to what you have said and what I have understood so far.
I am still slightly confused. Why do we displace(x) the object further as the question doesn't acquire you to do so? And if I have attached a 0.40Kg mass to the spring, will it not start oscillating with the given amplitude? Also, why does not the maximum potential energy occur at maximum extension upwards? At +A in this case (Third diagram in the image).
I apologise if I am being silly. Maybe, I am visualising the entire thing improperly. Thank you sir.
Posted from TSR Mobile
In questions like this you first place the mass on the spring.
The spring extends. The mass is then stationary there. In equilibrium. So there is no motion. Extension = e
If you want the mass to oscillate you have to displace it from its equilibrium position, so you extend it a bit more. (x)
The mass then oscillates about its original equilibrium position.
The question asks for the maximum PE stored in the spring. Not the PE of the mass.
PE stored in this spring depends on its extension. The bigger the extension, the more the PE.
The spring extends. The mass is then stationary there. In equilibrium. So there is no motion. Extension = e
If you want the mass to oscillate you have to displace it from its equilibrium position, so you extend it a bit more. (x)
The mass then oscillates about its original equilibrium position.
The question asks for the maximum PE stored in the spring. Not the PE of the mass.
PE stored in this spring depends on its extension. The bigger the extension, the more the PE.
Thank You very much Sir.
Posted from TSR Mobile
Posted from TSR Mobile
Original post by Stonebridge
In questions like this you first place the mass on the spring.
The spring extends. The mass is then stationary there. In equilibrium. So there is no motion. Extension = e
If you want the mass to oscillate you have to displace it from its equilibrium position, so you extend it a bit more. (x)
The mass then oscillates about its original equilibrium position.
The question asks for the maximum PE stored in the spring. Not the PE of the mass.
PE stored in this spring depends on its extension. The bigger the extension, the more the PE.
The spring extends. The mass is then stationary there. In equilibrium. So there is no motion. Extension = e
If you want the mass to oscillate you have to displace it from its equilibrium position, so you extend it a bit more. (x)
The mass then oscillates about its original equilibrium position.
The question asks for the maximum PE stored in the spring. Not the PE of the mass.
PE stored in this spring depends on its extension. The bigger the extension, the more the PE.
So in Summary, maximum potential energy of THE SPRING is;
at -A (maximum bottom amplitude) where (0.5)(k)((e+x)^2)
at +A (maximum highest amplitude) where (0.5)(k)((e-x)^2)
Maximum potential energy of THE MASS is: (0.5)(k)((A)^2)
Is this summary correct sir? Thank You.
Posted from TSR Mobile
Original post by 06awaism
So in Summary, maximum potential energy of THE SPRING is;
at -A (maximum bottom amplitude) where (0.5)(k)((e+x)^2)
at -A (maximum bottom amplitude) where (0.5)(k)((e+x)^2)
Yes
at +A (maximum highest amplitude) where (0.5)(k)((e-x)^2)
No. That's minimum pe of the spring.
I already stated that the maximum pe of the spring is at maximum extension.
Maximum potential energy of THE MASS is: (0.5)(k)((A)^2)
No.
Max pe of mass is at its maximum height. This is gravitational pe.
Posted from TSR Mobile
Thank you. I have made final notes. Could you have a brief look if you don't mind to see whether if they are correct or not.
The second question I would like to ask is related to momentum and impulse.
Statement 1) If the change in momentum = loss in momentum, is this represented by (-)magnitude ?
Statement 2) And if the change in momentum = gain in momentum, is this represented by (+)magnitude ?
For example in this image,
the answer to the question 4bíí) is = -8400. Is this value negative because of the statement 1?
Thank you very much sir.
Posted from TSR Mobile
The second question I would like to ask is related to momentum and impulse.
Statement 1) If the change in momentum = loss in momentum, is this represented by (-)magnitude ?
Statement 2) And if the change in momentum = gain in momentum, is this represented by (+)magnitude ?
For example in this image,
the answer to the question 4bíí) is = -8400. Is this value negative because of the statement 1?
Thank you very much sir.
Posted from TSR Mobile
The diagram on the right is not correct.
The diagram on the left at the bottom is correct for the elastic potential energy.
BUT
Gravitational potential energy is mgh. Correct.
The maximum gravitational potential of the mass is when h is a maximum.
This occurs at the maximum height of the mass which is at the top at A+ in your diagram. Not at the bottom.
Momentum.
If momentum (or any quantity) gets less, the change is negative.
It it get larger or greater the change is positive.
This is because
change = final value - initial value
The diagram on the left at the bottom is correct for the elastic potential energy.
BUT
Gravitational potential energy is mgh. Correct.
The maximum gravitational potential of the mass is when h is a maximum.
This occurs at the maximum height of the mass which is at the top at A+ in your diagram. Not at the bottom.
Momentum.
If momentum (or any quantity) gets less, the change is negative.
It it get larger or greater the change is positive.
This is because
change = final value - initial value
Thank you sir.
I have amended the diagrams in terms of the maximum height.
Is there anything else that needs changing?
Thank you again sir. I really appreciate your effort.
Posted from TSR Mobile
I have amended the diagrams in terms of the maximum height.
Is there anything else that needs changing?
Thank you again sir. I really appreciate your effort.
Posted from TSR Mobile
Original post by 06awaism
Thank you sir.
I have amended the diagrams in terms of the maximum height.
Is there anything else that needs changing?
Thank you again sir. I really appreciate your effort.
Posted from TSR Mobile
I have amended the diagrams in terms of the maximum height.
Is there anything else that needs changing?
Thank you again sir. I really appreciate your effort.
Posted from TSR Mobile
The diagram on the right is incorrect.
The formula is also incorrect.
Maximum elastic PE is ½k(e+x)2
not ½ke2
You have it correct under the left diagram, but incorrect under the right hand diagram.
The left diagram is the one to use.
Shall I just rub out the right diagram?
And every time a question asks me to find out the maximum potential energy of a spring, then I should always use (0.5)(k)((e+x)^2), which is the second bottom left diagram? Thank you.
Posted from TSR Mobile
And every time a question asks me to find out the maximum potential energy of a spring, then I should always use (0.5)(k)((e+x)^2), which is the second bottom left diagram? Thank you.
Posted from TSR Mobile
Original post by 06awaism
Shall I just rub out the right diagram?
And every time a question asks me to find out the maximum potential energy of a spring, then I should always use (0.5)(k)((e+x)^2), which is the second bottom left diagram? Thank you.
Posted from TSR Mobile
And every time a question asks me to find out the maximum potential energy of a spring, then I should always use (0.5)(k)((e+x)^2), which is the second bottom left diagram? Thank you.
Posted from TSR Mobile
Yes to both questions.
Thank You very much.
Posted from TSR Mobile
Posted from TSR Mobile
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