# Quantum Chemistry

Chapter05 - Home
Exercises: 1 - 2 - 3 - 4 - 5 - 6 - 7 - 8 - 9 - 10 - 11 - 12 - 13 - 14 - 15 - 16 - 17 - 18 - 19 - 20 - 21 - 22 - 23 - 24 - 25 - 26 - 27 - 28 - 29 - 30 - 31 - 32 - 33

Exercise. This is the solution to exercise 5.1 in the book.

Solution. We need to show that

$x\left(t\right)=Asin\omega t+Bcos\omega t$

verifies

$m\frac{{d}^{2}x}{d{t}^{2}}+kx=0$

We have

$\frac{{d}^{2}x}{d{t}^{2}}=-A{\omega }^{2}sin\omega t-B{\omega }^{2}cos\omega t$

Hence

$\frac{{d}^{2}x}{d{t}^{2}}=-{\omega }^{2}x$

If

$\omega =\sqrt{\frac{k}{m}}$

then

$m\frac{{d}^{2}x}{d{t}^{2}}+kx=0$