AQA physics paper On potential energy in mass spring system

[bonbon22]

Homework Statement

The spring constant of a helical spring is 28 N mñ1. A 0.40 kg mass is suspended from the

spring and set into simple harmonic motion of amplitude 60 mm.

Relevant Equations

1/2 F * change in extension = E
1/2 k * x^2 = E

The spring constant of a helical spring is 28 N mñ1. A 0.40 kg mass is suspended from the

spring and set into simple harmonic motion of amplitude 60 mm.
i use the equation 1/2 change in extension times by force where the force i assume is mass times by 9.8
but the mark scheme uses 1/2 kx ^2 and arrives at a different answer I am missing something but i don't know what it is?
the answer i got 0.392

answer in mark scheme 0.56

 

[Master1022]

The spring constant of a helical spring is 28 N mñ1. A 0.40 kg mass is suspended from the

spring and set into simple harmonic motion of amplitude 60 mm.
i use the equation 1/2 change in extension times by force where the force i assume is mass times by 9.8
but the mark scheme uses 1/2 kx ^2 and arrives at a different answer I am missing something but i don't know what it is?
the answer i got 0.392

answer in mark scheme 0.56

Amplitude is usually defined with respect to equilibrium. If this is a vertical system, then you will need to think about the extension required to reach equilibrium and include that in your EPE calculation. By this, I am talking about the spring extension required to balance the mass if it was stationary. So you will have the extension to equilibrium and the amplitude.

Hope that is of some help. Please reply if that doesn't yield the correct answer and I will have another look.

 

[bonbon22]

Amplitude is usually defined with respect to equilibrium. If this is a vertical system, then you will need to think about the extension required to reach equilibrium and include that in your EPE calculation. By this, I am talking about the spring extension required to balance the mass if it was stationary. So you will have the extension to equilibrium and the amplitude.

Hope that is of some help. Please reply if that doesn't yield the correct answer and I will have another look.

ahh yess i forgot to mention i did try that i found the extra static extention is 0.14 m including the amplitude
so when i plug it into the equation i use these values
0.4 * 9.8 * 1/2 * (0.14 + 60 *10 ^ -3 ) = 0.392

thank you for replying but i still do not quite understand

 

[Master1022]

ahh yess i forgot to mention i did try that i found the extra static extention is 0.14 m including the amplitude
so when i plug it into the equation i use these values
0.4 * 9.8 * 1/2 * (0.14 + 60 *10 ^ -3 ) = 0.392

thank you for replying but i still do not quite understand

Just to confirm, are we trying to find the EPE at maximum extension here? I will keep looking at this.

 

[bonbon22]

Just to confirm, are we trying to find the EPE at maximum extension here? I will keep looking at this.

the question phrased it as
the maximum potential energy stored in the spring during the first oscillation.

i should have mentioned that my bad

 

[Master1022]

ahh yess i forgot to mention i did try that i found the extra static extention is 0.14 m including the amplitude
so when i plug it into the equation i use these values
0.4 * 9.8 * 1/2 * (0.14 + 60 *10 ^ -3 ) = 0.392

thank you for replying but i still do not quite understand

I have just realized where I think you may have made the error. It is what you have done with the \frac{1}{2}Fx. That formula arises from the straight line graph of a static load and the resulting extension. However, this load is not being loaded statically and allowed to extend naturally (it is being set in motion), and thus this formula doesn't really work. Much better to think of work we do as \frac{1}{2} k x^2. Basically, we cannot use mg here as it does more work than just EPE in this scenario.

To understand why the work we do is \frac{1}{2}, think about \int \vec F \cdot d \vec x where F = kx. Happy to flesh this out for you if necessary if you haven't seen much calculus before.

On a side note, you could also just consider energy. So we know energy is conserved here. Therefore:
GPE lost by mass -------> KE of mass + EPE in spring, but at the bottom, there is no KE, so:
GPE lost by mass -------> EPE in spring
so you could also use mgh to get the answer as well.