Physics Problems & Solutions: Hooke's Law

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

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

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