91影视

The length of the allowed wave, L = 2x

The oscillators potential energy,$U\left(x\right)=\frac{1}{2}k{x}^2$

Standing wave condition is 2L/n.

The expression of kinetic energy is p²/2m.

The expression to calculate the energy of the state in infinite well is as follows.

$E=\frac{n^2{蟺}^2{\mathrm{鈩�}}^2}{2m{L}^2}$ 鈥︹�� (i)

Here, is the order of the state, is the reduced plank鈥檚 constant, is the mass of the particle and is the distance between the classical turning points.

The tunning point is produced when the potential energy is equation to the kinetic energy.

The kinetic energy is given by,

$U\left(x\right)=\frac{1}{2}k{t}^2$

Substitute E for U (x) into above equation.

$\mathrm{E}=\frac{1}{2}{\mathrm{kx}}^2$

$x^2=\frac{2E}{k}$

$x=\sqrt{\frac{2E}{k}}$

Therefore, the location of the x is localid="1660133237957" $\sqrt{\frac{2E}{k}}$.

The length of the allowed wave is given by, L = 2x

Substitute $\sqrt{\frac{2E}{k}}$ for localid="1658321346034" $x$ into the above equation

$\begin{array}{l}L=2\sqrt{\frac{2E}{k}}\\ L=\sqrt{\frac{8E}{k}}\end{array}$

The total energy is given by,

localid="1660133242572" $E=\frac{n^2{蟺}^2{\mathrm{鈩�}}^2}{2m{L}^2}+PE$

The kinetic energy is equal to potential energy for the turning point.

localid="1660133250120" $$\begin{gathered} E=\frac{n^2{蟺}^2{\mathrm{鈩�}}^2}{2m{L}^2}+\frac{n^2{蟺}^2{\mathrm{鈩�}}^2}{2m{L}^2} \\ E=\frac{n^2{蟺}^2{\mathrm{鈩�}}^2}{m{L}^2} \end{gathered}$$

Substitute $\sqrt{\frac{8E}{k}}$ for localid="1658321330233" $L$in equation (2).

$E=\frac{n^2{\mathrm{鈩�}}^2{蟺}^2}{2m\left(\frac{8E}{k}\right)}+P.E.$

Assume that kinetic energy is equal to potential energy.

$\begin{array}{l}E=\frac{n^2{\mathrm{鈩�}}^2{蟺}^2}{2m\left(\frac{8E}{k}\right)}+\frac{n^2{\mathrm{鈩�}}^2{蟺}^2}{2m\left(\frac{8E}{k}\right)}\\ E=\frac{2n^2{\mathrm{鈩�}}^2{蟺}^2}{2m\left(\frac{8E}{k}\right)}\end{array}$

Solve the above equation for localid="1658321323761" $E$.

localid="1660133256922" $\begin{array}{c}{E}^2=\frac{n^2{\mathrm{鈩�}}^2{蟺}^2}{m\left(\frac{8k}{k}\right)}\\ E^2=\frac{n^2{\mathrm{鈩�}}^2{蟺}^{2}k}{8m}\\ E=\left(\frac{蟺}{\sqrt{8}}\right)n\mathrm{鈩�}\sqrt{\frac{k}{m}}\end{array}$

Hence the energylocalid="1660133263314" $E=\left(\frac{蟺}{\sqrt{8}}\right)n\mathrm{鈩�}\sqrt{\frac{k}{m}}$is close to the harmonic oscillator鈥檚 energy and equal to space between the levels.