Question

Hello all,
Would someone kindly point me in the right direction for the following problem.

Two Hydrogen atoms oscillate symmetrically relative to their centre mass. Force between the two is:

\[F(x)=\frac{ 2U }{ b }(e ^{-2(x-x0)/b} - e ^{(x-x0)/b})\]

Motion of the atom is simple harmonic, thus the force between the two atoms behaves likes a spring and the two atoms act like two masses joined by a spring.

The frequency of oscillation is 1.31 x 10^14, mass of each atom is 1.67 x 10^-27.

Find the value of the spring constant.

Answer 1

could u plz write the expression more clearly?

Answer 2

\[m \ddot x+F(x)=0\]
But it is said that motion is simple harmonic,
if this is a case you can calculate k using this equation;
\[\omega=\sqrt{\frac{ k }{m }}=2\pi f \Rightarrow k=4\pi^2f^2m\]

Answer 3

I tried to solve that DE with that weird F(x) function but even wolfram can't do it so I think this simple harmonic solution is correct

Answer 4

Thanks so much Fifciol.

Apologies if the F(x) expression was not clear. The correct one is:

\[f(x) = \frac{ 2U }{ b } (e^{-2(x-x _{0})/b} - e ^{(x-x _{0})/b})\]

Answer 5

The simple harmonic solution looks correct but I feel im missing the point of the question. I cant solve that DE and am stuck at that point....