exam_1_review

The following questions and problems are courtesy of Justin Dunlap

*MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. *

Which of the following are characteristics of a mass in simple harmonic motion?

- The motion repeats at regular intervals.
- The motion is sinusoidal.
- The restoring force is proportional to the displacement from equilibrium.
- [….] A) 1 and 3 only
- [….] B) 2 and 3 only
- [….] C) 1 and 2 only
- [ X ] D) all of the above
- [….] E) none of the above

All of 1-3 are true and follow the definition (p. 417 in the text) of simple harmonic motion.

If the amplitude of the motion of a simple harmonic oscillator is doubled, by what factor does the frequency of the oscillator change?

- [….] A) 2
- [ X ] B) 1
- [….] C) 4
- [….] D) $\frac{1}{4}$
- [….] E) $\frac{1}{2}$

The new motion of the same oscillator system (assuming the same mass and the same spring constant, for example, or the same length and the same *g*) has the same frequency as the old one: $\;f_\text{new}=f_\text{old}\times 1\;$, therefore the answer is 1.

If your heart is beating at 76.0 beats per minute, what is the frequency of your heart's oscillations?

- [….] A) 2.54 Hz
- [….] B) 4560 Hz
- [….] C) 3.98 Hz
- [ X ] D) 1.27 Hz
- [….] E) 1450 Hz

The frequency in Hertz is the same as the number of cycles (beats) per second, 1/60^{th} of the beats per minute (since 1 min = 60 s):

- $f_\text{BPS}=\frac{76.0}{60}$= 1.27 Hz

A mass of 1.53 kg is attached to a spring and the system is undergoing simple harmonic oscillations with a frequency of 1.95 Hz and an amplitude of 7.50 cm. What is the total mechanical energy of the system?

- [….] A) 0.955 J
- [….] B) 0.633 J
- [ X ] C) 0.646 J
- [….] D) 0.844 J
- [….] E) 0 J

- Maximum value of potential energy of a mass on a spring:
- $U_\text{pot max}= \frac{1}{2}kx_\text{max}^2=E$, where $E$ is the total energy of the system. Note, when $U_\text{pot}$ is maximized, $U_\text{kin}\!=\!0$
- $x_\text{max}= 0.075\,$m
- $m=1.53\,$kg

- Find
*k*from the known frequency $f$, mass*m*and the spring constant*k*:- $f=\frac{\omega}{2\pi}$ $= \frac{1}{2\pi}\sqrt{\frac{k}{m}}$ $=1.95\,$Hz
- $f^2$ $= \frac{1}{4\pi^2}\!\cdot\!\frac{k}{m}$
- $k =4\pi^2f^2m$ $=4\!\cdot\!(3.14159)^2\!\cdot\!(1.95\,{\text{Hz}})^2\!\cdot\!1.53\,{\text{kg}}$ $\approx 230\frac{\text N}{\text m}$

- Plug this value of
*k*into $E$:- $E=\frac{1}{2}\big(4\pi^2f^2m\big)x_\text{max}^2$
- $E=\frac{1}{2}\!\cdot\!230\frac{\text N}{\text m}\!\cdot\!(0.075\,{\text{m}})^2$ = 0.646 J

A wave pulse traveling to the right along a thin cord reaches a discontinuity where the rope becomes thicker and heavier. What is the orientation of the reflected and transmitted pulses?

- [….] A) The reflected pulse returns right side up while the transmitted pulse is inverted.
- [….] B) Both are inverted.
- [….] C) Both are right side up.
- [ X ] D) The reflected pulse returns inverted while the transmitted pulse is right side up.
- [….] E) It is impossible to predict.

This is because a tight end would cause an inverted reflection and an open end would cause a right-side-up reflection. The transition to a heavier rope, in this example, is closer to the “tight end” condition: an infinitely heavy rope would be equivalent to the perfectly tight end. Therefore, the reflection will be inverted. The transmitted pulse is right-side-up in any case, because of vertical momentum conservation.

By what amount does the intensity level decrease when you triple your distance from a source of sound?

- [….] A) 12 dB
- [….] B) 4.8 dB
- [….] C) 3.0 dB
- [ X ] D) 9.5 dB
- [….] E) 6.0 dB

- In our 3-dimensional world, the energy of sound spreads out over the surface area of an expanding spherical wave-front, and therefore the intensity is inversely proportional to the square of the distance from the source.
- $I_2=\frac{1}{3^2}I_1$ $=\frac{I_1}{9}$

- The reduction in the intensity level is given by:
- $-\Delta\beta$ $=\beta_1-\beta_2$ $=10\log\frac{I_1}{I_\text{t.h.}}-10\log\frac{I_2}{I_\text{t.h.}}$ $=10\left(\log\frac{I_1}{I_\text{t.h.}}-\log\frac{I_2}{I_\text{t.h.}}\right)$ $=10\log\frac{\big(\frac{I_1}{I_\text{t.h.}}\big)}{\big(\frac{I_2}{I_\text{t.h.}}\big)}$ $=10\log\frac{I_1}{I_2}$ $=10\cdot\log 9$ $\approx 9.5\,$dB
- Make sure you can do this on your calculator (without confusing the decimal “log”, sometimes denoted “lg”, with the natural one “ln”)
- Note that the reference “threshold of hearing” intensity $I_\text{t.h.}=10^{-12}\frac{\text W}{\,{\text m}^2}\;$ drops out from this calculation and is not necessary.

An open pipe of length $L$ is resonating at its fundamental frequency. Which statement is correct?

- [….] A) The wavelength is $2L$ and there is a displacement antinode at the pipe's midpoint.
- [….] B) The wavelength is $\frac{3}{2}\!L$ and there are two displacement antinodes located inside the pipe.
- [….] C) The wavelength is $L$ and there is a displacement node at the pipe's midpoint.
- [….] D) The wavelength is $L$ and there is a displacement antinode at the pipe's midpoint.
- [ X ] E) The wavelength is $2L$ and there is a displacement node at the pipe's midpoint.

A pipe open at both ends has pressure nodes at those ends, which correspond to the displacement antinodes (the air is free to come and go at the open end, but the pressure is forced to equal 1 atm, so that $\Delta P=0$) The fundamental frequency of a resonator with identical ends corresponds to a half-wavelength that fits into the resonator of length *L*. Therefore the full wavelength is 2*L*. Since there are displacement antinodes at the ends, the midpoint should be the displacement node for the fundamental (because the midpoint is 1/4 wavelength away from the end… moving $\frac{1}{4}\!\lambda$ away from an antinode should land you at the node!)

A factory siren indicating the end of a shift has a frequency of 80 Hz. What frequency is perceived by the occupant of a car traveling away from the factory at 30 m/s? The speed of sound in air is 343 m/s.

- [….] A) 75 Hz
- [….] B) 81 Hz
- [….] C) 77 Hz
- [ X ] D) 73 Hz
- [….] E) 79 Hz

$f'=f\left(1-\frac{u}{v}\right)$ $=80\,{\text{Hz}}\cdot\left(1-\frac{30}{343}\right)$ $\approx 73\,$Hz. We use the minus sign in the Doppler formula because the distance between the source and the observer is increasing, “incorporating” some of the periods emitted by the source. Therefore the observer receives fewer periods than those that were emitted (i.e. the observed frequency is lower)

A fisherman fishing from a pier observes that the float on his line bobs up and down, taking 2.4 s to move from its highest to its lowest point. He also estimates that the distance between adjacent wave crests is 48 m. What is the speed of the waves going past the pier?

- [ X ] A) 10 m/s
- [….] B) 5.0 m/s
- [….] C) 20 m/s
- [….] D) 115 m/s
- […] E) 1.0 m/s

$T=2\cdot 2.4\,$s = 4.8 s. The speed of waves is $\;v=\frac{\lambda}{T}$ $=\frac{48\,{\text m}}{4.8\,{\text s}}=10\,\frac{\text m}{\text s}$

For an $xyz$-coordinate system shown to the right, if the *E*-vector is in the $+z$ direction, and the *B*-vector is in the $+x$ direction, what is the direction of propagation of the electromagnetic waves?

- [ X ] A) $+y$
- [….] B) $+x$
- [….] C) $+z$
- [….] D) $-x$
- [….] E) $-y$

The electromagnetic wave propagates in the direction of $\vec{\mathbf E}\times\vec{\mathbf B}$, that is in the $+y$ direction (in this right-handed system, $\hat{z}\times\hat{x}=\hat{y}$)

The energy density of an electromagnetic wave is

- [….] A) entirely in the magnetic field.
- [….] B) 1/4 in the electric field and 3/4 in the magnetic field.
- [….] C) 1/4 in the magnetic field and 3/4 in the electric field.
- [….] D) entirely in the electric field.
- [ X ] E) equally divided between the magnetic and the electric fields.

At any moment in time and in any point in space, electromagnetic radiation's energy density is always distributed equally between the magnetic energy density and the electric energy density components

Two objects are in all respects identical except for the fact that one was coated with a substance that is an excellent reflector of light while the other was coated with a substance that is a perfect absorber of light. You place both objects at the same distance from a powerful light source so they both receive the same amount of energy *U* from the light. The linear momentum these objects will receive is such that:

- [….] A) The reflecting object receives a smaller amount of momentum.
- [….] B) Both objects receive the same amount of momentum.
- [ X ] C) The reflecting objects receives a larger amount of momentum.
- [….] D) None of the previous answers is correct.

Just as with a ball bouncing off of a wall compared to the ball that sticks to the wall, the reflective surface receives twice the momentum compared to the absorptive case (the same amount to “stop” the light, plus another equal amount from the “recoil” of the reflected light re-emitted in the opposite direction)

A car is approaching a radio station at a speed of 25.0 m/s. If the radio station broadcasts at a frequency of 74.5 MHz, what change in frequency does the driver observe?

- [ X ] A) 6.21 Hz
- [….] B) 726 Hz
- [….] C) 64.5 Hz
- [….] D) 67.0 Hz
- [….] E) 98.3 Hz

$\Delta f=f'-f$ $=f\left(1+\frac{u}{c}\right)-f$ $=f\frac{u}{c}$ $=74.5\times 10^6\,{\text{Hz}}\cdot\frac{25\,\frac{\text m}{\text s}}{3\times 10^8\frac{\text m}{\text s}}$ $=6.21\,$Hz

A certain part of the electromagnetic spectrum ranges from 200 nm to 400 nm. What is the highest frequency associated with this portion of the spectrum?

- [ X ] A) $1.50 \times 10^{15}\,$Hz
- […] B) $7.50 \times 10^{14}\,$Hz
- […] C) $7.50 \times 10^{15}\,$Hz
- […] D) $7.50 \times 10^{13}\,$Hz
- […] E) $1.50 \times 10^{14}\,$Hz

The highest frequency corresponds to the shortest wavelength $\left(f=\frac{c}{\lambda}\right)$, therefore we need to convert $\lambda=200\,$nm to $f$ using this formula:

- $f=\frac{c}{\lambda}$ $=\frac{3\times 10^8\frac{\text m}{\text s}}{200\times 10^{-9}\,{\text m}}$ $=1.5\times 10^{15}\,$Hz

Which one of the following is the correct order of the electromagnetic spectrum from low to high frequencies?

- [ X ] A) radio waves, microwaves, infrared, visible, UV, X-rays, gamma rays
- […] B) radio waves, UV, X-rays, microwaves, infrared, visible, gamma rays
- […] C) radio waves, infrared, X-rays, microwaves, UV, visible, gamma rays
- […] D) radio waves, microwaves, visible, X-rays, infrared, UV, gamma rays
- […] E) radio waves, infrared, microwaves, UV, visible, X-rays, gamma rays

The visible should be in-between the UV and the infrared.

*Solve three of the four problems (cross out the one you do not want graded).
Show all of your work to receive full credit, most importantly show all the formulas you used to find the final answers. No credit will be awarded if an answer is given without work shown.*

Four waves are described by the following expression where distances are measured in meters and times in seconds.

- $\;\;\;y = 0.12 \cos\,(3x - 21t)$
- $\;\;\;y = 0.15 \sin\,(6x + 42t)$
- $\;\;\;y = -0.13 \cos\,(6x + 21t)$
- $\;\;\;y = -0.27 \sin\,(3x - 12t)$
- $\;\;\;y = 0.15 \sin\,(-9x - 36t)$

- a) Which of these waves travel in the $+x$ direction? (3 pts)
- waves 1 and 4
- (look for opposite signs in front of the
*x*and*t*terms inside the cos or sin function)

- b) Which of these waves have the same wavelength as wave 1? (3 pts)
- wave 4
- (look for the same magnitude of the coefficient in front of
*x*inside the cos or sin function)

- c) Which of these waves have the same amplitude as wave 2? (3 pts)
- wave 5
- (look for the same magnitude of the coefficient in front of the cos or sin function)

- d) Which of these waves have the same period as wave 3? (3 pts)
- wave 1
- (look for the same magnitude of the coefficient in front of
*t*inside the cos or sin function)

- e) Which of these waves have the same speed as wave 4? (3 pts)
- wave 5
- (look for the same magnitude of the ratio of the coefficients in front of
*t*and in front of*x*inside the cos or sin function)

See Equation sheet for Ch. 14, bullet #6 for details…

Wave | Direction | Amplitude (m) | Wavelength (m) | Period (s) | Frequency (Hz) | Magnitude of speed $\left(\frac{\mathbf m}{\mathbf s}\right)$ |
---|---|---|---|---|---|---|

1 | $+x$ | 0.12 | $\frac{2\pi}{3}$ | $\frac{2\pi}{21}$ | $\frac{21}{2\pi}$ | 7 |

2 | $-x$ | 0.15 | $\frac{\pi}{3}$ | $\frac{\pi}{21}$ | $\frac{21}{\pi}$ | 7 |

3 | $-x$ | 0.13 | $\frac{\pi}{3}$ | $\frac{2\pi}{21}$ | $\frac{21}{2\pi}$ | 3.5 |

4 | $+x$ | 0.27 | $\frac{2\pi}{3}$ | $\frac{\pi}{6}$ | $\frac{6}{\pi}$ | 4 |

5 | $-x$ | 0.15 | $\frac{2\pi}{9}$ | $\frac{\pi}{18}$ | $\frac{18}{\pi}$ | 4 |

A musician (Andrew Bird) used a double spinning horn speaker during a recent tour. While one horn spins toward you, the other spins away. Say the horns are emitting a frequency of 880 Hz, are spinning with an angular velocity of 2.0 rad/s, and that each horn is 1.0 m long.

- a) Due to a Doppler shift, what is the greatest pitch (frequency $f$) you will hear? Will this be when the horn is spinning towards you or away from you? (Remember, $v_t = A\omega$, where $A$ is the distance from the rotation axis to the opening of the horn) (3 pts)
- Greatest pitch - motion towards you
- $f'=f\left(\frac{1}{1-\frac{u}{v}}\right)$ $=f\left(\frac{1}{1-\frac{A\omega}{v}}\right)$ $=880\,{\text{Hz}}\,\left(\frac{1}{1-\frac{1.0\,{\text m}\,\cdot\,2.0\,\frac{\text{rad}}{\text s}}{340\,\frac{\text m}{\text s}}}\right)$ $=885\,$Hz

- b) What is the lowest pitch you will hear from the speakers due to a Doppler shift? Will this be when the horn is spinning towards you or away from you? (3 pts)
- Lowest pitch - motion away from you
- $f'=f\left(\frac{1}{1+\frac{u}{v}}\right)$ $=f\left(\frac{1}{1+\frac{A\omega}{v}}\right)$ $=880\,{\text{Hz}}\,\left(\frac{1}{1+\frac{1.0\,{\text m}\,\cdot\,2.0\,\frac{\text{rad}}{\text s}}{340\,\frac{\text m}{\text s}}}\right)$ $=875\,$Hz

- c) What is the beat frequency you hear from this instrument? (3 pts)
- $f_\text{beat}=\big|\,f_1-f_2\big|$ $=\big|885\,{\text{Hz}}-875\,{\text{Hz}}\big|$ $=10\,$Hz

- d) At the concert your friend sitting 1 m from the speakers hears an intensity of $9.0\times 10^{-2}\frac{\text W}{\,{\text m}^2}$. What intensity do you hear sitting 10 m away from the speakers? (3 pts)
- $I=\frac{\text{Power}}{4\pi r^2}$
- $\frac{I_\text{friend}}{I_\text{you}}=\frac{(10\,{\text m})^2}{(1\,{\text m})^2}$ $=100$
- $I_\text{you}=\frac{1}{100}I_\text{friend}$ $=9.0\times 10^{-4}\frac{\text W}{\,{\text m}^2}$

- e) What is the intensity level in decibels that your friend hears? (3 pts)
- $\beta=10\,{\text{dB}}\log\left(\frac{I_1}{I_\text{t.h.}}\right)$ $=10\,{\text{dB}}\log\left(\!\frac{9.0\times 10^{-2}\frac{\text W}{\,{\text m}^2}}{ 10^{-12}\frac{\text W}{\,{\text m}^2}}\!\right)$ $\approx 110\,$dB

Match the definitions and descriptions with the best term or phrase given below (1.5 pts each):

Resonance Underdamping Critical damping Overdamping Period Frequency Angular frequency Wavelength Simple pendulum Physical pendulum Transverse wave Longitudinal wave Linear mass density | Wave function Reflections Ultrasound Intensity Decibel Doppler effect Superposition Constructive interference Destructive interference Standing wave Fundamental frequency 2 ^{nd} Harmonic frequency 3 ^{rd} Harmonic frequency | Beat frequency In-phase sources Opposite phase sources Direction of propagation of light Speed of light in a vacuum Microwaves Radio waves Infrared Visible Ultraviolet x-rays Gamma rays |

- Two sources that emit crests at the same time and emit troughs at the same time.
- In-phase sources

- Ninety degrees to both the electric field and the magnetic field in a moving electromagnetic wave.
- Direction of propagation of light

- Energy per unit time per unit area.
- Intensity

- A measure of sound intensity. Increasing by about three of these units will indicate a doubling of the intensity of the sound.
- Decibel

- When two waves combine to create a wave with an amplitude less than either of the two original waves.
- Destructive interference

- A part of the electromagnetic spectrum with frequencies just greater than those visible by humans.
- Ultraviolet

- A point mass on the end of a mass-less string that is allowed to swing back and forth.
- Simple pendulum

- A wave where the direction of the molecules is perpendicular to the direction of the wave.
- Transverse wave

- The lowest frequency that can be created on a string or in a tube.
- Fundamental frequency

- The length of time between two wave crests.
- Period

The air pressure variations in a sound wave cause the eardrum (tympanic membrane) to vibrate.

- For a given vibration amplitude, are the maximum velocity and acceleration of the eardrum greatest for high frequency sounds or low frequency sounds? (1 pts)
- $v_\text{max}$ $=x_\text{max}\omega$
- $a_\text{max}$ $=x_\text{max}\omega^2$
- the greatest values are for the high-frequency sound

- Find the maximum velocity and the maximum acceleration of the eardrum for vibrations of amplitude $1.0\times 10^{-8}\,$m at a frequency of 20.0 kHz. (5 pts)
- $v_\text{max}$ $=x_\text{max}\omega$ $=x_\text{max}(2\pi f)$ $=1.0\times 10^{-8}\,$m$\,\cdot\,6.283\cdot 20\times 10^3\,$Hz $=1.26\times 10^{-3}\frac{\text m}{\text s}$
- $a_\text{max}$ $=x_\text{max}\omega^2$ $=x_\text{max}(2\pi f)^2$ $=1.0\times 10^{-8}\,$m$\,\cdot\,(6.283\cdot 20\times 10^3\,$Hz$)^2=158\,\frac{\text m}{\,{\text s}^2}$

- What is the period of a complete oscillation of the ear drum at this frequency? (2 pts)
- $T=\frac{1}{f}$ $=\frac{1}{20\times 10^3\,{\text{Hz}}}$ $=5.0\times 10^{-5}\,$s

- Using a crude model of the eardrum as a mass (3.0 mg) on a spring, what would be the spring constant of the eardrum, assuming the resonance frequency of 20.0 kHz? (3 pts)
- $T=2\pi\sqrt{\frac{m}{k}}$
- $\left(\frac{T}{2\pi}\right)^2=\frac{m}{k}$
- $k=\frac{4\pi^2m}{T^2}$ $=\frac{4\,\cdot\,3.14159^2\,\cdot\,3.0\times 10^{-6}\,{\text{kg}}}{\big(5.0\times 10^{-5}\,{\text s}\big)^2}$ $=4.74\times 10^4\frac{\text N}{\text m}$

- The ear canal (external auditory canal) can be modeled as a tube with one closed end. If the length of the ear canal is 25 mm long and the speed of sound in air is 340 m/s, what is the fundamental (1
^{st}harmonic) of the ear canal? (4 pts)- $f_1=\frac{v}{4L}$ $=\frac{340\,\frac{\text m}{\text s}}{4\,\cdot\,25\times 10^{-3}\,{\text m}}$ $=3400\,$Hz

exam_1_review.txt · Last modified: 2014/04/19 19:26 by wikimanager

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