Physics Problems & Solutions: Harmonic Oscillator

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Classical Mechanics - Harmonic Oscillator

A particle of mass m undergoes harmonic oscillation with period T₀. A force f proportional to the speed v of the particle, f = −bv, is introduced. If the particle continues to oscillate, the period with f acting is

A. Larger than T₀
B. Smaller than T₀
C. Independent of b
D. Dependent linearly on b
E. Constantly changing

(GR9677 #08)

Solution:

f = −bv → minus sign means f is restoring force (damped oscillation).
The oscillation is getting slower (larger period) before it finally comes to stop.

Answer: A

Quantum Mechanics - Probability

A system is know to be in the normalized state described by the wave function

$\psi(\theta,\varphi)=\frac{1}{\sqrt{30}}(5Y_4^3+Y_6^3-2Y_6^0)$

Where $Y_l^m (\theta,\varphi)$ are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m = 3 is

A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15

(GR9277 #28)

Solution:

For the state $\left | \alpha\right \rangle = \sum_{i}c_i\left | a^{(i)}\right \rangle$ , probability is $P= \sum_{i}\left |c_i \right |^2$

States with m = 3 → Y_4^3

and

with c₁ = 5/√30 and c₂ = 1/√30

Therefore,

$P= \left |\frac{5}{\sqrt{30}} \right |^2+ \left |\frac{1}{\sqrt{30}} \right |^2=\frac{26}{30}=\frac{13}{15}$

Answer: E

Quantum Mechanics - Harmonic Oscillator

If v is frequency and h is Planck’s constant, the ground state energy of a one-dimensional quantum mechanical harmonic oscillator

A. 0
B. 1/3 hv
C. 1/2 hv
D. hv
E. 3/2 hv

(GR9277 #56)

Solution:

The energy of a 1-D quantum HO: E_n = (n + ½)hv

The ground state energy, n = 0 → E₀ = ½ hv

Answer: C

Classical Mechanics - Physical Pendulum

A long, straight, and massless rod pivots about one end in a vertical plane. In configuration I, shown above, two small identical masses are attached to the free end; in configuration II, one mass is moved to the center of the rod. What is the ratio of the frequency of small oscillations of configuration II to that of configuration I?

A. (6/5)^½
B. (3/2)^½
C. 6/5
D. 3/2
E. 5/3

(GR9277 #61)

Solution:

Angular frequency for physical pendulum:

ω = (MgL/I)^1/2

Pendulum I
M = 2m
I = mr² + mr² = 2mr²
L = r + r = 2r

ω_I= (2mg2r/2mr²)^1/2= (2g/r)^1/2

Pendulum II
M = 2m
I = mr² + m(r/2)² = (⁵/₄)mr²

L = ½ r + r = (³/₂)r

ω_II=[ 2mg (³/₂)r / (⁵/₄)mr²]^1/2= (12g/5r)^1/2

The ratio:

ω_II/ ω_I= (12g/5r)^1/2/ (2g/r)^1/2= 6/5

Answer: A

Electromagnetism - Coulomb's Law

Two point charges with the same charge +Q are fixed along the x-axis and are a distance 2R apart as shown. A small particle with mass m and charge –q is placed at the midpoint between them. What is the angular frequency ω of small oscillations of this particle along the y-direction?

A. $\frac{Qq}{2\pi \epsilon_\theta mR^2}$

B. $\frac{Qq}{4\pi \epsilon_\theta mR^2}$

C. $\frac{Qq}{2\pi \epsilon_\theta mR^3}$

D. $\left (\frac{Qq}{4\pi \epsilon_\theta mR^2} \right )^{\frac{1}{2}}$

E. $\left (\frac{Qq}{2\pi \epsilon_\theta mR^3} \right )^{\frac{1}{2}}$

(GR9277 #65)

Solution:

Check the unit, ω → sec⁻¹

with
Q = q → Coulomb
R = meter
m = kg
ϵ₀ = coulomb² / (newton · meter²) = coulomb²·sec²/ (kg · meter³)

A. coulomb² · kg · meter³/ (coulomb² · sec² · kg · meter²) = meter/sec²
→ FALSE.

B. Identical unit with A. → FALSE

C. 1/sec² → FALSE

D. √m/sec → FALSE

E. 1/sec → TRUE

Answer: E

Calculation

Coloumb's Law: $F=k\frac{Qq}{r^2}$

In this case,

$\vec{F}=\frac{1}{4\pi \epsilon_0}\frac{2Q(-q)}{R^2} \text{ sin} \theta \text{ }\hat j =-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \text{ sin} \theta\text{ }\hat j$

(Electric fields from both Qs are in x-and y-direction, but in x-axis, they cancel each other.)

For small θ, sin θ ≈ tan θ = y/R

Hooke's Law in y-axis: $F=m\ddot{y}=-ky$

Angular frequency, $\omega = \sqrt{k/m} = - \ddot{y}/y\rightarrow \ddot{y} = -\omega^2 y$

$\begin{aligned} m\ddot{y}&=-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \frac{y}{R} \\ -m\omega^2 y &=-\frac{1}{2\pi \epsilon_0}\frac{Qq}{R^2} \frac{y}{R} \\\omega^2 &= \frac{Qq}{2\pi \epsilon_0mR^3} \\\omega &= \left ( \frac{Qq}{2\pi \epsilon_0mR^3} \right )^{\frac{1}{2}} \end{aligned}$

Quantum Mechanics - Harmonic Oscillator

The energy levels for the one-dimensional harmonic oscillator are hν(n + ½), n = 0,1,2,⋯ How will the energy levels for the potential shown in the graph above differ from those for the harmonic oscillator?

A. The term ¹⁄₂ will be changed to ³⁄₂
B. The energy of each level will be doubled.
C. The energy of each level will be halved.
D. Only those for even values of n will be present.
E. Only those for odd values of n will be present.

(GR9277 #89)

Solution:

See GR9677 #98

Answer: E

Classical Mechanics - Wave equation

The equation $y=A \sin2\pi\left(\frac{1}{T}-\frac{x}{\lambda}\right)$ where A, T, and λ are positive constants, represents a wave whose

A. Amplitude is 2A
B. Velocity is in the negative x–direction
C. Period is T/λ
D. Speed is x/t
E. Speed is λ/T

(GR8677 #04)

Solution:

(A) FALSE.
Amplitude = maximum displacement = y_max = A.

(B) FALSE.
For a wave traveling to the right (positive x-direction): y = ƒ(x − vt)
(x − vt) = constant.
As t increases, x must increases to keep (x − vt) = constant

For a wave traveling to the left (negative x-direction): y = ƒ(x + vt)
As t increases, x decreases to keep (x + vt) = constant.

The problem gives: y = ƒ(vt − x)
As t increases, x must increases to keep (x − vt) = constant.
The waves is traveling to the right.

(C) FALSE.
The unit of T/λ (second/meter) does not match with the unit is period (second).

(D) FALSE.
The speed of the wave is λ/T.

(E) TRUE.
See D.

Answer: E

Classical Mechanics - Harmonic Oscillator

A particle of mass m that moves along the x-axis has potential energy V(x) = a + bx² , where a and b are positive constants. Its initial velocity is v₀ at x = 0. It will execute simple harmonic motion with a frequency determined by the value of

A. b alone
B. b and a alone
C. b and m alone
D. b, a and m alone
E. b, a, m and v₀

(GR8677 #60)

Solution:

$\\F=-\nabla V=-\frac{\partial V}{\partial x}=-\frac{\partial}{\partial x}(a+bx^2)=-2bx \\F=-kx \\\rightarrow -2bx=-kx\rightarrow k=2b \\\omega=\sqrt{k/m}=2\pi f\rightarrow k =4\pi^2f^2 m \\4\pi^2f^2 m=2b\rightarrow f=\sqrt{\frac{2b}{4\pi^2m}}$

Answer: C

Classical Mechanics - Hooke's Law

A particle is constrains to move along the x-axis under the influence of the net force F = − kx with amplitude A and frequency f, where k is a positive constant. When x = A/2, the particle speed is

A. 2πfA
B. √3πfA
C. √2πfA
D. πfA
E. (1/3) πfA

(GR8677 #77)

Solution:

$F=-kx=m\ddot{x}\rightarrow m\ddot{x}+kx=0$

Equation of motion: $\ddot{x}+\omega^2x=0$
with angular velocity: ω = √(k/m) = 2πf

Solution to the equation of motion, wave function: x = A sin ωt
Velocity: v = dx/dt = Aω cos ωt

x = A/2 → A sin ωt = A/2
sin ωt = 1/2 → ωt = 30^o
cos 30^o = ½√3

v = Aω cos ωt = A 2πf ½√3 = √3πfA

Answer: B

Classical Mechanics - Pendulum

Which of the following best illustrates the acceleration of a pendulum bob at points a through e?

(GR0177 #01)

Solution:

The acceleration of pendulum: a = a_centripetal + a_tangential
a_cent = v²/r = ω²r
a_tan = αr

At equilibrium (position B): ω = constant
α = dω/dt = 0
a = a_cent

At maximum amplitude (position A and C): v = 0
a = a_tan

At other positions:
a = a_cent + a_tan

Answer: C

Thermal Physics - Equipartition Law

A three-dimensional harmonic oscillator is in thermal equilibrium with a temperature reservoir at temperature T. The average total energy of oscillator is

A. ½kT
B. kT
C. ³⁄₂kT
D. 3kT
E. 6kT

(GR0177 #05)

Solution:

Equipartition of Energy: E = ½ fkT

f = Degree of Freedom (DoF)

3D harmonic oscillator has 6 DoF = 3 components of momentum (kinetic energy) and 3 components of position (potential energy)
→ E = ⁶⁄₂ kT = 3kT

Answer: D

Quantum Mechanics - Harmonic Oscillator

Let $|n\rangle$ represent the normalized n^th energy eigenstate of the one-dimensional harmonic oscillator,

$H|n\rangle=\hbar\omega\left(n+\frac{1}{2}\right)|n\rangle$

If $|\psi\rangle$ is a normalized ensemble state that can be expanded as a linear combination

$|\psi\rangle=\frac{1}{\sqrt{14}}|1\rangle-\frac{2}{\sqrt{14}}|2\rangle+\frac{3}{\sqrt{14}}|3\rangle$

of the eigenstates, what is the expectation value of the energy operator in this ensemble state?

A. $\frac{101}{14} \hbar \omega$

B. $\frac{43}{14} \hbar \omega$

C. $\frac{23}{14} \hbar \omega$

D. $\frac{17}{\sqrt{14}} \hbar \omega$

E. $\frac{7}{\sqrt{14}} \hbar \omega$

(GR0177 #45)

Solution:

The energy eigenstates:

$\begin{aligned} H|1\rangle&=\hbar\omega\left(1+\frac{1}{2}\right)|1\rangle = \frac{3}{2}\hbar\omega|1\rangle \\H|2\rangle&=\frac{5}{2}\hbar\omega|2\rangle \\H|3\rangle&=\frac{7}{2}\hbar\omega|3\rangle \end{aligned}$

The expectation value:

$\begin{aligned} \langle H\rangle &=\langle\psi|H|\psi\rangle \\&=\frac{3}{2}\hbar\omega\left( \frac{1}{\sqrt{14}}\right )^2\langle1|1\rangle+\frac{5}{2}\hbar\omega\left(- \frac{2}{\sqrt{14}}\right )^2\langle2|2\rangle+\frac{7}{2}\hbar\omega\left( \frac{3}{\sqrt{14}}\right )^2\langle3|3\rangle \\&=\left( \frac{3}{28} +\frac{20}{28}+\frac{63}{28}\right)\hbar\omega \\&=\frac{86}{28}\hbar\omega =\frac{43}{14}\hbar\omega \end{aligned}$

Answer: B

Electromagnetism - LC circuit

For an inductor and capacitor connected in series, the equation describing the motion of charge is

$L \frac{d^2 Q}{dt^2}+\frac{1}{C} Q=0$

where L is the inductance, C is capacitance, and Q is the charge. An analogous equation can be written for a simple harmonic oscillator with position x, mass m, and spring constant k. Which of the following correctly lists the mechanical analogs of L, C, and Q?

	L	C	Q
A	m	k	x
B.	m	1/k	x
C.	k	x	m
D.	1/k	1/m	x
E.	x	1/k	1/m

(GR0177 #59)

Solution:

The form of SHO: $m\ddot{x}+kx=0$

LC circuit equation: $L\ddot{Q}+\frac{Q}{C}=0$

L = m
C = 1/k
Q = x

Answer: B

Classical Mechanics - Harmonic Oscillator

Two identical blocks are connected by a spring. The combination is suspended, at rest, from a string attached to the ceiling, as shown in the figure. The string breaks suddenly. Immediately after the string breaks, what is the downward acceleration of the upper block?

A. 0
B. g/2
C. g
D. √2 g
E. 2g

(GR0177 #72)

Solution:

In equilibrium (before the string is cut):
Block 1:

$\sum F_1 = 0 \rightarrow T-m_1g-kx=0$ …(Eq.1)