A linear harmonic oscillator with mass m and frequency w is in the stationary state with energy 5/2homega What is the standard (root-mean-square)deviation of the oscillator momentum in this state? Enter the ratio of the found standard deviation to (h)/(l)(wherel=sqrt ((h)/(momega ))) in the answer box after rounding it to the second decimal place. ouranswer

To find the standard deviation of the oscillator momentum in the given state, we can use the Heisenberg uncertainty principle. The uncertainty principle states that the product of the uncertainties in position and momentum is greater than or equal to a constant.<br /><br />The energy of the linear harmonic oscillator is given by:<br /><br />$E = \frac{1}{2}m\omega x^2 + \frac{p^2}{2m}$<br /><br />where $x$ is the position and $p$ is the momentum.<br /><br />In the stationary state with energy $5/2h\omega$, the average value of the momentum squared is:<br /><br />$\langle p^2 \rangle = \frac{h\omega}{2m}$<br /><br />The standard deviation of the momentum is given by:<br /><br />$\sigma_p = \sqrt{\langle p^2 \rangle - \left(\frac{\langle p \rangle}{2}\right)^2}$<br /><br />Since the oscillator is in a stationary state, the average momentum $\langle p \rangle$ is zero. Therefore, the standard deviation of the momentum is:<br /><br />$\sigma_p = \sqrt{\frac{h\omega}{2m}}$<br /><br />The ratio of the found standard deviation to $\frac{h}{l}$ is:<br /><br />$\frac{\sigma_p}{h/l} = \frac{\sqrt{\frac{h\omega}{2m}}}{h/l} = \frac{\sqrt{2}}{2}$<br /><br />Rounding to the second decimal place, the answer is:<br /><br />$\frac{\sigma_p}{h/l} = 0.71$