Collection of Solved Problems in Physics

Macroscopic oscillator

Task number: 4415

• Hint to 1) and 2)

  How can the energy of the oscillator be determined from the given values in classical physics? And what does the energy of a linear harmonic oscillator (LHO) look like from the perspective of quantum mechanics?

• Solution to hint

  The energy of the system, at the moment it passes through the equilibrium position, is equal to the kinetic energy of the oscillating body, i.e., \[E=\frac{1}{2} m v^2 ,\] where the mass \(m\) and velocity \(v\) are given in the task assignment. The potential energy of the spring is zero at this instant.

   

  From the perspective of quantum mechanics, the LHO has a discrete energy spectrum given by

  \[E_n=\left(n+\frac{1}{2}\right)\,\hbar \omega=\left(n+\frac{1}{2}\right)\,h f ,\hspace{20px}n = 0, 1, 2, ... ,\]

  where the oscillation frequency \(f\) is known from the task assignment.

• Solution 1

  Both the classical and quantum mechanical descriptions refer to the same system with the given energy. Therefore, we can equate the classical and quantum expressions

  \[\left(n+\frac{1}{2}\right)\,h f = \frac{1}{2}mv^2 ,\]

  from which

  \[n=\frac{mv^2}{2hf}\,-\,\frac{1}{2} \, ,\]

  and after substituting the numerical values

  \[n=\frac{(1{\cdot} 10^{-6}\,\mbox{kg})\,\cdot\, (0{,}1\,\mbox{m · s}^{-1})^2}{2\,\cdot \,(6{,}626{\cdot} 10^{-34}\,\mbox{J · s})\,\cdot \,(1\,\mbox{Hz})}\,-\,\frac{1}{2}\,\dot{=}\,\ 7{,}55\,\cdot\,10^{24}\approx 10^{25} .\]

• Solution 2

  From the quantum mechanical perspective, the LHO has a discrete energy spectrum given by

  \[ E_n=\left(n+\frac{1}{2}\right)\,\hbar \omega=\left(n+\frac{1}{2}\right)\,h f,\ \hspace{20px}n = 0,\,1,\,2,\,... \ , \]

  where the oscillation frequency \(f\) is known from the task assignment.

  The energy difference between two adjacent levels is therefore constant

  \[ \Delta E=hf=6{,}626\,\cdot\,10^{-34}\ \mbox{J} . \]

  The ratio of this difference to the total energy of the oscillator is

  \[ \frac{\Delta E}{E}=\frac{hf}{\frac{1}{2}mv^2}=\frac{2\,\cdot\,6{,}626\,\cdot\,10^{-34}}{1\,\cdot\,10^{-6}\,\cdot\,(0{,}1)^2}=13{,}252\,\cdot\,10^{-26}\approx 10^{-25}. \]

  Thus, the total energy is roughly 10²⁵ times larger than the energy difference between two neighbouring levels, which corresponds to the quantum number calculated in Solution 1.

• Hint to 3)

  How does the speed of the oscillator change if its position varies according to \(x=A\sin(\omega t)\)?

• Solution to hint

  In this case, the velocity is given by the derivative of the position

  \[v=\frac{\mbox{d}x}{\mbox{d}t}=A\omega\cos(\omega t) .\]

• Solution 3

  If the position of the oscillator varies according to

  \[ x(t)=A\sin(\omega t) , \]

  its velocity changes according to

  \[ v(t)=\frac{\mbox{d}x}{\mbox{d}t}=A\omega\cos(\omega t) . \]

  At times \(t=k\pi\) the particle passes through the equilibrium position and moves (as stated in the task assignment) with a velocity of \(v\) = 10 cm·s⁻¹. Therefore,

  \[ v=A\omega=A\,2\pi f , \]

  from which the amplitude is

  \[ A=\frac{v}{2\pi f}=\frac{0{,}1\,\mbox{m · s}^{-1}}{2\pi\,\cdot\,1\,\mbox{Hz} }\,\dot{=}\,\,0{,}016\ \mbox{m}=16\ \mbox{mm} . \]

   

  We now focus on the ground state of such an oscillator from the quantum mechanical perspective. Its energy is given by \(E_1=hf\). As it passes through the equilibrium position, it must again hold that

  \[ E_1=\frac{1}{2}mv^2 , \] \[ hf=\frac{1}{2}mA^2 \omega^2=\frac{1}{2}mA^2 (2\pi f)^2=2mA^2 \pi^2 f^2, \]

  from which the amplitude is

  \[ A=\sqrt{\frac{h}{2\pi^2 mf}}=\sqrt{\frac{6{,}626\,\cdot\,10^{-36}}{2\pi^2\,\cdot\,10^{-6}}}\ \mbox{m}\,\dot{=}\,0{,}58\,\cdot\,10^{-15}\ \mbox{m} , \]

  which is roughly ten times smaller than the typical diameter of an atomic nucleus.

• Answer

  1) Although the described body with a mass of 1 mg appears very small to the naked eye, from a physical standpoint it is a macroscopic object. Consequently, it is unsurprising that the quantum number describing its state as a quantum LHO is approximately 7,55·10²⁴, an enormous value.

   

  2) The energy difference between two adjacent levels is constant \(\Delta E = 6{,}626 \cdot 10^{-34} J\).

  The ratio of this difference to the total energy of the oscillator is of the order of \[\frac{\Delta E}{E}\approx 10^{-25} .\] This indicates that, on such scales, energy quantisation loses practical significance, as it is entirely unobservable.

   

  3) The amplitude of oscillation for the given parameters is approximately 16 mm. For the ground state of the oscillator (i.e., the state with quantum number \(n=1\)), this amplitude would be roughly 0,58·10⁻¹⁵ m, about ten times smaller than the typical diameter of an atomic nucleus.