Physics Problems & Solutions: #28

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Electromagnetism - Oscilloscope

The figure above represents the trace on the screen of cathode ray oscilloscope. The screen is graduated in centimeters. The spot on the screen moves horizontally with a constant speed of 0.5 centimeter/millisecond and the vertical scale is 2 volts/centimeter. The signal is a superposition of two oscillations. Which of the following are most nearly the observed amplitude and frequency of these two oscillations?

	Oscillation 1	Oscillation 2
A.	5V, 250Hz	2.5V, 1000Hz
B.	1.5V, 250Hz	3V, 1500Hz
C.	5V, 6Hz	2V, 2Hz
D.	2.5V, 83Hz	1.25V, 500Hz
E.	6.14V, 98Hz	1.35V, 257Hz

(GR9677 #28)

Solution:

The graph shows one big λ consists of 6 small λs.
λ_big≈ 6 cm
λ_small≈ 1 cm

v = λf

Given: v = 0.5 cm/ms
f_big= v/λ_b = 0.5 cm/(6 cm ms) = 1/ (12 ms) = 10³/(12 s) = 83 Hz (Osc. 1)
f_small = v/λ_s = 0.5 cm/(1 cm ms) = 1/ (2 ms) = 10³/(2 s) = 500 Hz (Osc. 2)

Answer: D

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 - Normalization of Wavefunction

Eigenfunction for a rigid dumbbell rotating about its center have a φ dependence of the form ψ(φ) = Ae^imφ, where m is a quantum number and A is a constant. Which of the following values of A will properly normalize the eigenfunction?

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

(GR8677 #28)

Solution:
Normalized Wavefunction: $\int |\psi(x)|^2dx=\int \psi(x)\psi^*(x)dx=1$
$\\\psi(\phi)=Ae^{im\phi} \\\psi^*(\phi)=Ae^{-im\phi}$

$\\\int_{0}^{2\pi} |\psi(\phi)|^2d\phi=\int_{0}^{2\pi} |A|^2d\phi=1 \\|A|^2\int_{0}^{2\pi}d\phi=|A|^2 2\pi=1\rightarrow A = \frac{1}{\sqrt{2\pi}}$

Answer: D

Quantum Mechanics- Wave Function

A. 10
B. 5
C. 0
D. −5
E. −10

(GR0177 #28)