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

#### The CdePster

1. The problem statement, all variables and given/known data
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.

2. Relevant 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 dont know how to get the period from the frequency then, well sadly you shouldnt be helping me lol :)!

3. 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 dont 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 dont believe im wrong but please enlighten me if I am. Thank you!!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### rock.freak667

Homework Helper
0.5kx2 gives the potential energy, 0.5kA2 gives the total energy. What value did you use for 'x'?

#### The CdePster

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 dont think I have surely I cant 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.

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