Physics Problems & Solutions: Potential Energy

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Quantum Mechanics - Potential Wall

Consider a potential of the form

V(x) = 0, x ≤ a
V(x) = V_0,a < x < b
V(x) = 0, x ≥ b

As shown in the figure above. Which of the following wave functions is possible for a particle incident from the left with energy E < V₀.

(GR9677 #18)

Solution:

A. Classic not QM potential → FALSE
B. No decrease in amplitude → FALSE
C. Decay exponentially inside the wall, decrease amplitude (fits E < V₀) → TRUE

D. QM Oscillator harmonics → FALSE
E. Cosine wave function not QM potential → FALSE

Answer: C

Classical Mechanics - Lagrangian

A particle of mass m on the Earth’s surface is confined to move on the parabolic curve y = ax², where y is up. Which of the following is a Lagrangian for the particle?

A. $L = \frac{1}{2} m\dot{y} ^2 \left (1+\frac{1}{4ay} \right )-mgy$

B. $L = \frac{1}{2} m\dot{y} ^2 \left (1-\frac{1}{4ay} \right )-mgy$

C. $L = \frac{1}{2} m\dot{x} ^2 \left (1+\frac{1}{4ay} \right )-mgx$

D. $L = \frac{1}{2} m\dot{x} ^2 \left (1+4a^2x^2 \right )+mgx$

E. $L = \frac{1}{2} m\dot{x} ^2 +\frac{1}{2} m\dot{y} ^2+mgy$

(GR9277 #44)

Solution:

Kinetic Energy, $T = \frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)$

Potential Energy,

Lagrangian, $L=T-U=\frac{1}{2} m(\dot{x} ^2 +\dot{y} ^2)-mgy$

Given the curve y = ax²,

$\begin{aligned} \dot{y}&=\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}=2ax\dot{x}\\\rightarrow \dot{x}&=\frac{\dot{y}}{2ax} \\\dot{x}^2&=\frac{\dot{y}^2}{4a^2x^2}=\frac{\dot{y}^2}{4ay} \end{aligned}$

$\begin{aligned} L&=\frac{1}{2} m\left (\frac{\dot{y} ^2}{4ay}+\dot{y}^2 \right ) -mgy \\&=\frac{1}{2} m\dot{y} ^2\left (\frac{1}{4ay}+1 \right ) -mgy \end{aligned}$

Answer: A

Classical Mechanics - Circular Motion

A particle of mass M moves in a circular orbit of radius r around a fixed point under the influence of an attractive force F = ^K⁄_r_³, where K is a constant. If the potential energy of the particle is zero at an infinite distance from the force center, the total energy of the particle in the circular orbit is

A. −^K⁄_r²
B. −^K⁄_2r²
C. 0
D. ^K⁄_2r²
E. ^K⁄_r²

(GR9277 #87)

Solution:

Attractive force = Centripetal Force
^K⁄_r_³ = ^mv²⁄_r
mv² = ^K⁄_r²

Kinetic energy: T = ½mv² = ^K⁄_2r²
Potential energy: V(r) = − ∫ F dr = −K ∫ ¹⁄_r_³ dr = ^K⁄_2r²
Attractive force → negative potential energy, V(r) = − ^K⁄_2r²
Total energy: T + V = ^K⁄_2r² − ^K⁄_2r² = 0

Answer: C

Quantum Mechanics - Schrodinger Equation

The wave function ψ(x) = A exp (−b²x²/2), where A and b are real constants, is a normalized eigenfunction of the Schrodinger equation for a particle of mass M and energy E in a one dimensional potential V(x) such that V(x) = 0 at x = 0. Which of the following is correct?

A. V = ħ²b⁴/2M
B. V = ħ²b⁴x²/2M
C. V = ħ²b⁶x⁴/2M
D. E = ħ²b²(1 − b²x²)
E. E = ħ²b⁴/2M

(GR8677 #18)

Solution:

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)\psi(x)=E\psi(x)$

$\begin{aligned} \frac{\partial\psi}{\partial x}&=\frac{\partial}{\partial x}e^{-b^2x^2/2}=-b^2x\psi \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2\frac{\partial}{\partial x}x\psi=-b^2\left( x\frac{\partial\psi}{\partial x}+\psi\frac{\partial x}{\partial x}\right ) \\\frac{\partial^{2}\psi}{\partial x^{2}}&=-b^2(-b^2x^2\psi+\psi)=(b^4x^2-b^2)\psi\end{aligned}$

Schrodinger Equation: $-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)\psi(x)+V(x)\psi(x)=E\psi(x)$

To find E → V(x = 0) = 0:

$\\-\frac{\hbar^{2}}{2m}(0-b^2)+0=E \\\rightarrow E=\frac{\hbar^{2}b^2}{2m}$

To find V(x):
$\\-\frac{\hbar^{2}}{2m}(b^4x^2-b^2)+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+\frac{\hbar^{2}b^2}{2m}+V(x)=E \\\\-\frac{\hbar^{2}b^4x^2}{2m}+E+V(x)=E \\\\\rightarrow V(x)=\frac{\hbar^{2}b^4x^2}{2m}$

Answer: B

Classical Mechanics - Conservative Force

A. ½mv²
B. −4kx³
C. kx⁴
D. −kx⁵/5
E. mg

(GR8677 #34)

Solution:

$F=-\nabla U =-\frac{\partial U}{\partial x} = -\frac{\partial}{\partial x}kx^4=-4kx^3$

Answer: B

Classical Mechanics - Hamiltonian

A. (p²/2m) + kx⁴
B. (p²/2m) − kx⁴
C. kx⁴
D. ½mv² − kx⁴
E. ½mv²

(GR8677 #35)

Solution:

Hamiltonian: H = T + U
U = kx⁴
T = ½mv² = p²/2m
H = (p²/2m) + kx⁴

Answer: A

Classical Mechanics - Lagrangian and Hamiltonian

Question 34-36

The potential energy of a body constrained to move on a straight line is kx⁴ where k is a constant. The position of the body is x, its speed v, its linear momentum p, and its mass m.

The body moves from x₁ at time t₁ to x₂ at time t₂. Which of the following quantities is an extremum for the x-t curve corresponding to this motion, if end points are fixed?

A. $\int_{t_1}^{t_2}\bigg(\frac{1}{2}mv^2-kx^4\bigg)dt$
B. $\int_{t_1}^{t_2}\frac{1}{2}mv^2dt$
C. $\int_{t_1}^{t_2}mxv$ $dt$
D. $\int_{x_1}^{x_2}\bigg(\frac{1}{2}mv^2+kx^4\bigg)dx$
E. $\int_{x_1}^{x_2}mv$ $dx$

(GR8677 #36)

Solution:

$\\$Lagrangian: $L=T-U \\$Hamiltonian: $ H=\int_{t_1}^{t_2} L dt \\U=kx^4 \\\\T=\frac{1}{2} mv^2 \\\rightarrow H=\int_{t_1}^{t_2}\bigg(\frac{1}{2} mv^2-kx^4\bigg)dt$

Answer: A

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

Electromagnetism - Energy

A system consists of two charged particles of equal mass. Initially the particles are far apart, have zero potential energy, and one particle has nonzero speed. If radiation is neglected, which of the following is true of the total energy of the system?

It zero and remains zero
It is negative and constant
It is positive and constant
It is constant, but the sign cannot be determined unless the initial velocities of both particles are known.
It cannot be a constant of the motion because the particles exert force on each other.

(GR8677 #78)

Solution:

Particles have zero potential energy → PE = 0

One particle has nonzero speed → v

0 → KE

0

→ Total energy: E = PE + KE = 0 + (KE

0) → E

0

No radiation → no energy loss → Energy constant.

→ Energy is positive and constant.

Answer: C

Quantum Mechanics - Infinite Potential Well

The figure above shows one of the possible energy eigenfunctions ψ(x) for a particle bouncing freely back and forth along the x-axis between impenetrable walls located at x = −a and x = +a. The potential energy equals zero for |x| > a. If the energy of the particle is 2 electron volts when it is in the quantum state associated with this eigenfunction, what is its energy when it is in the quantum state of lowest possible energy?

A. 0 eV
B. 1/√2 eV
C. 1/2 eV
D. 1 eV
E. 2 eV

(GR8677 #90)

Solution:

Impenetrable walls = infinite potential walls.
The initial wavefunctions for the first four states in the system:

So,

represents n = 2.

The energies for infinite potential walls: E_n = n² E₁

Since E₂ = 2 eV → 2² E₁ = 2 → E₁ = 2/4 = 1/2 eV

Answer: C

Quantum Mechanics - Free Particle

A free particle with initial kinetic energy E and de Broglie wavelength λ enters a region in which it has potential energy V. What is the particle’s new de Broglie wavelength?

A. λ(1 + E/V)
B. λ(1 − V/E)
C. λ(1− E/V)^{− 1}
D. λ(1 + V/E)^½
E. λ(1 − V/E)^{− ½}

(GR0177 #46)

Solution:

The initial kinetic energy of free particle:
$E=\frac{p^2}{2m}\rightarrow p=\sqrt{2mE}$

De Broglie wavelength:
$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Energy of the particle when it enters the region:

So, its wavelength becomes:
$\lambda_f=\frac{h}{\sqrt{2mE_f}}=\frac{h}{\sqrt{2m(E-V)}}$

$\\\lambda_f^2=\frac{h^2}{2mE}\bigg(\frac{E}{E-V}\bigg)=\frac{h^2}{2mE}\bigg(\frac{1}{1-V/E}\bigg)\\\\\\\lambda_f=\frac{h}{\sqrt{2mE}}\bigg(\frac{1}{1-V/E}\bigg)^{1/2}=\lambda(1-V/E)^{-1/2}$

Answer: E

Classical Mechanics - Potential Energy

A Particle of mass m moves in a one-dimensional potential V(x) = −ax² + bx⁴, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to

A. π(a/2b)^1/2
B. π(a/m)^1/2
C. (a/mb)^1/2
D. 2(a/m)^1/2
E. (a/2m)^1/2

(GR9677 #92)

Solution:

The minima of the potential (most probable value x or the equilibrium position of the mass):

$\frac{dV(x)}{dx}=0$

$\begin{aligned} \frac{dV(x)}{dx}&=\frac{d}{dx} (-ax^2+bx^4)\\&=-2ax+4bx^3\\&=x(4bx^2-2a)=0 \\ \\x_1&=0 \\ x_2&=\sqrt{a/2b} \end{aligned}$

Angular Frequency: $\omega=\sqrt{\frac{k}{m}}$

Conservative force:
$\\F=-\nabla{V} =-kx \rightarrow \frac{dF}{dx} = k = \frac{d^2V(x)}{dx^2}$

$\begin{aligned} k=\frac{d}{dx}(-2ax+4bx^3)&=-2a+12bx^2 \\k(x=0)&= -2a \\k(x=\sqrt{a/2b})&=-2a+6a= 4a \end{aligned}$

Thus, the angular frequency about the minima of the potential:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{4a}{m}}=2\sqrt{\frac{a}{m}}$

Answer: D

Classical Mechanics - Period of the Motion

A particle of mass m moves in the potential above. The period of the motion when the particle has energy E is

A. $\sqrt{k/m}$

B. $2\pi\sqrt{m/k}$

C. $2\pi\sqrt{2E/mg^2}$

D. $\pi\sqrt{m/k}+2\sqrt{2E/mg^2}$

E. $2\pi\sqrt{m/k}+4$

(GR9677 #93)

Solution:

Total Period: T = T_SHO+ T_grav

For V = ½kx²→ Simple Harmonic Oscillator (SHO)
Period of SHO, T_SHO= 2π√(k/m)

The graph shows only half of the usual SHO potential:
T_SHO= ½ × 2π√(k/m) = π√(k/m)

Total Period: T = π√(k/m) + T_grav

Answer: D

Notes:

To find T_grav with V = mgx:

E = T + V = 0 + mgx
x = E/mg

Kinematic Equation:

x = v₀t + ½gt²
v₀= 0
x = ½gt²= E/mg
t²= 2E/mg²
t = T_grav= √(2E/mg²)

Since the particle has to travel from the origin to the right endpoint and then back to the origin, the total time contribution from this potential:

T_grav= 2√(2E/mg²)

The total period is, T = π√(k/m) + 2√(2E/mg²)

Quantum Mechanics - Harmonic Oscillator

A particle of mass m is acted on by a harmonic force with potential energy function V(x) = mω²x²/2 (a one dimensional simple harmonic oscillator). If there is a wall at x = 0 so that V = ∞ for x < 0, then the energy levels are equals to

A. 0, ħω, 2ħω, ...
B. 0, ¹⁄₂ħω, ħω, ...
C. ¹⁄₂ħω, ³⁄₂ħω, ⁵⁄₂ħω, ...
D. ³⁄₂ħω, ⁷⁄₂ħω, ¹¹⁄₂ħω, ...
E. 0, ³⁄₂ħω, ⁵⁄₂ħω, ...

(GR9677 #98)

Solution

The probability distributions for the quantum states of the oscillator without the barrier.

image: hyperphysics

Infinite barrier at the origin means a node at origin, or the wave function goes to zero at x = 0.

By symmetry, the ground state will disappear, as well all the even states. Odd values remain.

Answer: D

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