Physics Problems & Solutions: #45

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Electromagnetism - High and Low-Pass Filters

The circuits below consist of two-element combinations of capacitors, diodes and resistors. V_inrepresents an ac-voltage with variable frequency. It is desired to build a circuit for which V_out≈ V_in at high frequencies and V_out≈ 0 at low frequencies. Which of the following circuits will perform this task?

(GR9677 #45)

Solution:

High-pass filter: V_out≈ V_in at high frequencies, (ω → ∞)
Low-pass filter: V_out≈ 0 at low frequencies (ω → 0)

Diode is a device to allow current flow in one direction only. A combination of Diode and Resistor will not affect the output frequency.
→ (B), (C) are FALSE

(A), (D), (E) circuits are combination of resistor and capacitor.

Capacitive reactance, X_C = 1 / ωC
X_C → 0 for ω → ∞
X_C → ∞ for ω → 0

(A) FALSE
It is parallel combination of capacitor and resistor, voltage is uniform throughout the circuit.

(D) FALSE
It is Low-pass filter only.
Using Voltage Divider for series circuit: V_out= X_C/(R + X_C) V_in

X_C → 0 for ω → ∞

V_out= 0/(R + 0) V_in≈ 0

(E) TRUE
It is High and Low-pass filter.
Using Voltage Divider: V_out= R/(R + X_C) V_in

X_C → 0 for ω → ∞

V_out= R/(R + 0) V_in
V_out≈ V_in

X_C → ∞ for ω → 0
V_out= R/(R + ∞) V_in
V_out≈ 0

Answer: E

Notes: for similar problem see GR0177 #39

Classical Mechanics - Energy

A ball is dropped from a height h. As it bounces off the floor, its speed is 80% of what it was just before it hit the floor. The ball will then rise to a height of most nearly

A. 0.94 h
B. 0.80 h
C. 0.75 h
D. 0.64 h
E. 0.50 h

(GR9277 #45)

Solution:

Conservation of energy before and when the ball hits the ground:

$\begin{aligned} \frac{1}{2}m \cdot 0+mgh&=\frac{1}{2}mv^2+mg\cdot 0 \\mh&=\frac{1}{2}v^2 \\v^2 &= 2gh \end{aligned}$

Conservation of energy when and after the ball hits the ground

$\begin{aligned} \frac{1}{2}mv'^2+mg\cdot 0&=\frac{1}{2}m\cdot 0+mgh' \\\frac{1}{2}mv'^2&=mgh' \\v'^2 &= 2gh' \\(0.8v)^2&=0.64v^2 = 0.64\times 2gh = 2gh' \\h'&=0.64h \end{aligned}$

Answer: D

Nuclear & Particle Physics - Photon Scattering

Photons of wavelength λ scatter elastically on free protons initially at rest. The wavelength of the photons scattered at 90^o is increased by

λ ⁄ 137
λ ⁄ 1836
ħ ⁄ (m_ec) where ħ Planck's constant, m_e the rest mass of an electron, and c the speed of light
ħ ⁄ (m_pc) where ħ Planck's constant, m_p the rest mass of a proton, and c the speed of light
zero

(GR8677 #45)

Solution:

Compton Scattering: the inelastic scattering of photons from charged particles.

$\lambda_f-\lambda_i=\Delta \lambda=\frac{h}{m_e c} (1- \cos \theta)$

θ = 90^o → cos θ = 0
For proton scattered at 90^o, Δλ = ħ ⁄ (m_pc)

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