Physics Problems & Solutions: #93

Showing posts with label #93. Show all posts

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)

Classical Mechanics - Pendulum

The figure above shown a small mass connected to a string, which is attached to a vertical post. If the mass is released when the string is horizontal as shown, the magnitude of the total acceleration of the mass as a function of the angle θ is

A. $g \sin \theta$
B. $2g \cos \theta$
C. $2g \sin \theta$
D. $g\sqrt{3\cos^2\theta+1}$
E. $g\sqrt{3\sin^2\theta+1}$

(GR9277 #93)

Solution:

Total acceleration:

$\vec{a}=\vec{a}_{\text{tangential}}+\vec{a}_{\text{centripetal}}$

At initial position where $\theta=0$ , there is no string tension $\rightarrow \vec{a}_{\text{centripetal}}=0$

$\vec{a}=\vec{a}_{\text{tangential}}+0=\vec{a}_{\text{tangential}}=g$

Using method of elimination:

A. $g \sin \theta = g \sin 0=0 \rightarrow$ FALSE

B. $2g \cos \theta=2g \cos 0=2g \rightarrow$ FALSE

C. $2g \sin \theta=2g \sin 0=0 \rightarrow$ FALSE

D. $g\sqrt{3\cos^2\theta+1}=g\sqrt{3(1)+1}=2g \rightarrow$ FALSE

E. $g\sqrt{3\sin^2\theta+1}=g\sqrt{0+1}=g \rightarrow$ TRUE

Answer: E

Classical Mechanics - Efficiency

An electric sander has a continuous belt that rubs against a wood surface as shown schematically above. The sander is 100 percent efficient and draws current of 9 amperes from a 120-volt line. The belt speed is 10 meters per second. If the sander is pushing against the wood with a normal force of 100 newtons, the coefficient of friction is most nearly

A. 0.02
B. 0.2
C. 0.4
D. 1.1
E. 10

(GR8677 #93)

Solution:

Efficiency:

$\eta =\frac{W_\text{done}}{W_\text{input}}=\frac{P_\text{done}}{P_\text{input}}=1$

$P_\text{done}=P_\text{input}$

$\\P_\text{done}=Fv=\mu N v=\mu(100)(10)=1000\mu\\ P_\text{input}= VI=(9)(120)=1080$

$1080=1000\mu \rightarrow \mu=1.08$

Answer: D

Quantum Mechanics - Bohr Radius

The solution to the Schrodinger equation for the ground state of hydrogen is

$\psi_0=\frac{1}{\sqrt{\pi a^3_0}}e^{-r/a_0}$

where a₀ is the Bohr radius and r is the distance from the origin. Which of the following is the most probable value for r?

A. 0
B. a₀ / 2
C. a₀
D. 2a₀
E. ∞

(GR0177 #93)

Solution:

Probability

$\begin{aligned} P(r)dr &=|\psi_0|^2dV\\&=|\psi_0|^2d \left(\frac{4}{3}\pi r^3 \right)\\&=|\psi_0|^2 4\pi r^2 dr\\&=\frac{4 r^2}{a^3_0} e^{-2r/a_0}dr \end{aligned}$

The most probable value of r corresponds to the peak of the plot of P(r) versus r.
The slope of the curve at this point is zero.

$\frac{dP(r)}{dr}=0$

$\begin{aligned} \frac{dP(r)}{dr} &=\frac{d}{dr}\frac{4 r^2}{a^3_0} e^{-2r/a_0} \\&=\frac{4}{a^3_0}\frac{d}{dr}r^2e^{-2r/a_0}\\&=\frac{4}{a^3_0}\left(2re^{-2r/a_0}+ \frac{-2}{a_0}e^{-2r/a_0}r^2\right)\\&=\frac{8 r}{a^3_0}e^{-2r/a_0}\left(1-\frac{r}{a_0} \right ) \end{aligned}$

For $\frac{dP(r)}{dr}=0$

$1-\frac{r}{a_0}=0\rightarrow r = a_0$

r = a₀

Answer: C