Calculating Period and Speed of Atomic Vibrations | Atom Motion in Solids

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Atoms in solids vibrate due to thermal energy, with typical amplitudes around 10^-9 cm and frequencies near 10^12 Hz. The period of oscillation can be calculated using the formula T = 1/f, resulting in approximately 10^-12 seconds. To find the maximum speed of an atom, the relationship between angular frequency (omega), amplitude (A), and mass (m) is used, where omega = sqrt(k/m) and maximum speed is given by v = omega x A. The discussion highlights the importance of careful calculations to avoid errors. Ultimately, the participants successfully derive the correct values for period and speed.
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Homework Statement


Atoms in a solid are in continuous vibrational motion due to thermal energy. At room temperature, the amplitude of these atomic bibrations is typically about 10^-9 cm, and their frequency is on the order of 10^12 Hz. What is the approximate period of oscillation of a typical atom? What is the mazimum speed of such an atom?


Homework Equations


(1/f)=T


The Attempt at a Solution


For period of oscillation i know the answer is 10^-12, but I do not know the formula for finding the speed of the atom. I used C=(lambda/f) but this didn't work. Any help would be appreciated. Thanks
 
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The energy of a particle in SHM is mv^2/2, where v is max speed and also kA^2/2, where A is amplitude. Do you know omega in terms of k and m? Then you can figure it out.
 
ok so omega = sqrt*k/m then multiply omega x A to get max speed? But i get a very small number and I know it's not correct, what am I doing wrong?
 
Last edited:
(omega = 2*pi*f = 2*pi/T.)

Probably doing some arithmetical errors. Show your calculations.
 
oh sweet, i got it thanks!
 

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