Energy in a spring system and static extension

[nokia8650]

The spring constant of a helical spring is 28Nm^-1. A 0.40 kg mass is suspended from the
spring and set into simple harmonic motion of amplitude 60mm.

Calculate
(i) the static extension produced by the 0.40 kg mass,

(ii) the maximum potential energy stored in the spring during the first oscillation.

The markscheme is shown below:

http://img407.imageshack.us/img407/980/38156466aq9.th.jpg


Id the question, and get the correct answer to both parts. However, I do not understand the second method of working out the second part of the question - why does one have to add "mgA" in the final part? I presume it is due to E = mgh, however do not understand the reasining behind the fact that it must be added - when the mass is at the equilibrium position, i thought all of its energy is a) kinetic energy, and b) eleastic potential energy.

Thanks

 

[kbaumen]

The spring constant of a helical spring is 28Nm^-1. A 0.40 kg mass is suspended from the
spring and set into simple harmonic motion of amplitude 60mm.

Calculate
(i) the static extension produced by the 0.40 kg mass,

(ii) the maximum potential energy stored in the spring during the first oscillation.

The markscheme is shown below:

http://img407.imageshack.us/img407/980/38156466aq9.th.jpg


Id the question, and get the correct answer to both parts. However, I do not understand the second method of working out the second part of the question - why does one have to add "mgA" in the final part? I presume it is due to E = mgh, however do not understand the reasining behind the fact that it must be added - when the mass is at the equilibrium position, i thought all of its energy is a) kinetic energy, and b) eleastic potential energy.

Thanks

The object has a potential energy difference between equilibrium and the lowest point. If the reference point is in fact the lowest point, at the lowest point the potential energy is W = 0 but at the equilibrium it is W = mgA, where A is amplitude or distance between equilibrium and the lowest point. If, while at equilibrium, we cut the spring, the object will start a free-fall and, if we don't take into account friction between the object and air (or the medium in which the oscillations occured), the potential energy, which it had at equilibrium, will have turned to kinetic energy at the point, where once was the lowest point of these oscillations.

I hope that helps.

 

[thomate1]

for your help - also, do you know if there are any good websites to learn about simple harmonic motion?
Hello,

I'm happy to help you understand the second part of the question. Let's break it down step by step.

In simple harmonic motion, the potential energy stored in the spring is given by the equation:

E = 1/2 * k * x^2

where E is the potential energy, k is the spring constant, and x is the displacement from the equilibrium position.

In part (i) of the question, we calculated the static extension, which is the displacement from the equilibrium position when the mass is suspended. This is given by 60mm or 0.06m. So, plugging in the values, we get:

E = 1/2 * (28 Nm^-1) * (0.06m)^2 = 0.0504 J

This is the total potential energy stored in the spring when the mass is at the equilibrium position.

Now, in part (ii) of the question, we are asked to calculate the maximum potential energy stored in the spring during the first oscillation. This means that we need to find the maximum displacement of the mass from the equilibrium position, which is the amplitude of motion. In this case, it is given as 60mm or 0.06m.

So, using the same equation as before, we get:

E = 1/2 * (28 Nm^-1) * (0.06m)^2 = 0.0504 J

This is the same answer as before, but it is not the maximum potential energy stored in the spring during the first oscillation. This is because when the mass is at the maximum displacement, it has both kinetic energy and potential energy. The potential energy stored in the spring is only a part of the total energy.

To find the maximum potential energy stored in the spring, we need to subtract the kinetic energy of the mass from the total energy. The kinetic energy of the mass is given by:

KE = 1/2 * m * v^2

where m is the mass and v is the velocity. In this case, the mass is 0.40 kg and the velocity at the maximum displacement is 0 m/s (since the mass is momentarily at rest before changing direction). So, the kinetic energy is 0 J.

Therefore, the maximum potential