SHM in atoms question I think my textbook is wrong :0 yet again.

AI Thread Summary
The discussion centers around calculating the total energy of atomic oscillations using the formula total energy = 0.5(mass)(angular frequency)^2 x max displacement. The user believes their calculations, which yield a total energy of 2.1E-11 J, are correct, while the textbook states the answer is 1.2E-20 J. They detail their method for finding angular frequency and emphasize that using amplitude as the maximum displacement should yield consistent results. The conversation highlights confusion over the significant discrepancy in the calculated energy values. The user seeks clarification on their approach and the potential error in the textbook.
The CdePster
Messages
9
Reaction score
0

Homework Statement


In a simple atomic model of a solid, the atoms vibrate with a frequency of 2E-11 Hz. The amplitude of the vibration of the atoms is 5.5E-10 m and the mass of each atom is 4.8E-26. Calculate the total energy of the oscillations of an atom.


Homework Equations


total energy= 0.5(mass)(angular frequency)^2 x max displacement
not forgetting angular frequency= 2pi/the period
Other information is easy and if you don't know how to get the period from the frequency then, well sadly you shouldn't be helping me lol :)!


The Attempt at a Solution


I solved it, its very easy you just plug in the numbers really however the answer given is 1.2E-20. I don't get this, I used the equation I gave you and also the easier equations like E=0.5kx^2 after finding the spring constant of the atoms from the equation T= 2pi root(m/k)
and I get the same answer either way which is 2.1E-11 J.
I don't believe I am wrong but please enlighten me if I am. Thank you!
 
Physics news on Phys.org
0.5kx2 gives the potential energy, 0.5kA2 gives the total energy. What value did you use for 'x'?
 
thanks for the reply, i used the amplitude of oscillation 5.5E-10 for 'x' or rather 'x0' As it would be in this case. Because x0 is the same as amplitude yes? Even if I had used the wrong value for x which i don't think I have surely I can't be one billion out as the answer would suggest?

Just to show you what I did so you know 100% that my method is OK. Using the first equation which I mentioned in the first post.
Total energy = 0.5*4.8E-26*(1.26E12)2*5.5E-10

And the way I got the angular frequency was by doing 1/2E11=5E-12
and then doing 2pi/5E-12 to give me 1.26E12.

This method is faster than using the E=0.5kx2 and T=2pi sqroot(m/k) and is actually the same thing because angular frequency2 = k/m.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top