For every question, also consider as a possible answer
E) none of the abovePossibly useful information:
v = x / tp = m v T = 2
a = v / tPE = m g h T = 2
v = vi + a t PE = (1/2) k x2v =
x = xi + vi t + (1/2) a t2KE = (1/2) m v2v=(wavelength) x (frequency)
v = r F = k x L = (n) x (half wavelength)
F = m a Ei = Ef
F12 = - F21pi = pf
w = mg F = p / t
g = 9.8 m/s2 10 m/s2
For every question, also consider as a possible answer
E) none of the above
1. Increasing the amplitude of a mass-and-spring simple oscillator makes its period
A) longer2. Increasing the spring constant k (that is, using a stronger spring) in a mass-and-spring simple harmonic oscillator makes its period
B) shorter
C) unchanged
A) longer3. Increasing the mass of a mass-and-spring simple harmonic oscillator makes its period
B) shorter
C) unchanged
A) longer4. A mass-and spring simple harmonic oscillator has maximum potential energy
B) shorter
C) unchanged
A) at its equilibrium position5. A mass-and spring simple harmonic oscillator has maximum kinetic energy
B) when its displacement equals its amplitude
C) half way between equilibrium and amplitude
D) two-thirds of the way between equilibrium and amplitude
A) at its equilibrium position6. The amplitude of a simple harmonic oscillator is
B) when its displacement equals its amplitude
C) half way between equilibrium and amplitude
D) two-thirds of the way between equilibrium and amplitude
A) the time required for one oscillation7. The frequency of a certain oscillator is 100 Hz; its period is
B) the number of oscillators per second
C) the energy stored in the oscillations
D) the maximum distance moved from equilibrium
A) 0.1 s8. There are "signals" of many different frequencies coming into the antenna of your radio. Only the one with a particular frequency is amplified and produces the sound you listen to. This is an example of
B) 0.01s; T = 1/f = 1/100 = 0.01 s
C) 0.001 s
D) 0.0001 s
A) resonance9. Which of the following is a longitudinal wave?
B) damping
C) timbre or quality
D) amplitude degeneration
A) wave on a string10. A wave has a frequency of 20 Hz and travels 5 m in one second. It has
B) light
C) sound
D) all of the above
A) a wave speed of 100 m/s and a wavelength of 4 m.11. A wave has a frequency of 32 Hz and travels 4 m in one second. It has
B) a wave speed of 100 m/s and a wavelength of 1/4 m.
C) a wave speed of 5 m/s and a wavelength of 1/4 mv = (freq) x (wavelength)D) a wave speed of 5 m and a wavelength of 4 m
wavelength = v / freq = (5 m/s) / (20 Hz) = (1/4) m
A) a wave speed of 100 m/s and a wavelength of 128 m.12. A wave has a frequency of 50 Hz and a wavelength of 0.5 m. It has a wave speed of
B) a wave speed of 100 m/s and a wavelength of 1/8 m.
C) a wave speed of 4 m/s and a wavelength of 1/8 mv = (freq) x (wavelength)D) a wave speed of 4 m and a wavelength of 8 m
wavelength = v / freq = (4 m/s) / (32 Hz) = (1/8) m
A) 2.5 m/s13. For standing waves on a string,
B) 10 m/s
C) 25 m/sv = (freq) x (wavelength)D) 100 m/s
v = (50 Hz) (0.5 m) = 25 m/s
A) a node is located at each end14. For standing waves on a string, the distance between adjacent nodes is always
B) a whole number times half the wavelength equals the length of the string
C) the whole "pattern" of standing waves occurs only for certain frequencies
D) all of the above
A) the length of the string15. For standing waves on a string, the distance between adjacent antinodes is always
B) one-half the wavelength
C) one wavelength
D) all of the above
A) half the length of the string16. A bobber on a fishing line oscillates up and down two (2) times per second as waves pass by. The waves have a wavelength of 25 cm. The waves are traveling at
B) one-half the wavelength
C) one wavelength
D) all of the above
A) 12.5 cm/s17. If you put your fingertip in a pool of water and repeatedly move it up and down, you will create circular water waves that move out from that point. What will happen to the wavelength of these waves if you move your finger up and down more slowly (or less frequently)?
B) 25 cm/s
C) 50 cm/sv = (freq) x (wavelength)D) 100 cm/s
v = (2 Hz) (25 cm) = 50 m/s
A) increase18. Sound isv = (wavelength) x (frequency)B) remain the sameA decrease in frequency means an increase in wavelength.
C) decrease
A) an electromagnetic wave19. "Infrasonic" meansB) a polarized wavelight is an electromagnetic wave
radio and television and X-rays are also electromagnetic waveC) a longitudinal waveonly transverse waves can be polarizedlongitudinal waves can not be polarized
D) all of the above
A) lower than the range of human hearing20. Bats and dolphins use echolocation to navigate or the find food or to find their way without relying on sight. The frequencies they use are
B) higher than the range of human hearing
C) faster than the speed of sound
D) slower than the speed of sound
A) supersonic21. The range of human hearing is about
B) infrasonic
C) ultrasonicD) microsonicBoth bats and dolphins use ultrasound with frequencies of about 50 kHz and above.
A) 10 Hz to 100 Hz22. Ella Fitzgerald made commercials for Memorex in which she used her voice to break a wine glass. This is an example of
B) 50 Hz to 500 Hz
C) 50 Hz to 20 000 Hz
D) 1 000 Hz to 100 000 Hz
A) echolocation23. Beats are heard when two sounds have
B) reflected sound
C) ultrasonic frequencies
D) resonance
A) nearly the same amplitude24. The fundamental frequency present in a sound is the
B) nearly the same frequencies
C) twice the amplitude
D) exactly twice the wavelength
A) sum of all the frequencies mixed together25. The fundamental frequency present in a sound determines the
B) difference between the highest and lowest frequencies present
C) lowest frequency present
D) highest frequency present
A) quality or timbre26. The "pitch" of a sound is determined by its
B) amplitude
C) pitch or note
D) all of the above
A) overtones frequencies27. The quality or timbre -- the distincitive characteristic -- of a sound is determined by its
B) harmonics frequencies
C) fundamental frequency
D) resonance frequencies
A) overtones or harmonics28. Consider a musical note of 440 hertz ( 'A' on the staff). Two octaves lower is represented by a musical note of
B) amplitude or loudness
C) attack or decay
D) frequency
A) 110 Hz29. Suppose you play a note of a certain pitch on a violin. You can produce a higher-pitched note by
B) 440 Hz
C) 660 Hz
D) 880 Hz
A) shortening the length of the string that is allowed to vibrate30. When a flute sound is viewed on an oscilloscope, the sound wave is very smooth. This is because
B) decreasing the tension of the string (loosening the string)
C) increasing the linear mass density of the string (using a "heavier" string)
D) lengthening the part of the string that vibrates.
A) the amplitude is always small (flutes are quiet)31. When a trumpet sound is viewed on an oscilloscope, the sound wave is very complex. This is because
B) it has practically no overtones or harmonics
C) its fundamental frequency has a smaller amplitude than its second and third harmonics
D) its harmonics get larger and larger.
A) the amplitude is always large (trumpets are loud)32. Increasing the mass of a simple pendulum makes its period
B) it has practically no overtones or harmonics
C) it has many overtones or harmonics
D) it has only even-numbered overtones or harmonics.
A) longer33. Increasing the length of a a simple pendulum makes its period
B) shorter
C) unchanged
A) longer34. The period of a simple pendulum depends upon its
B) shorter
C) unchanged
A) mass35. Ordinary household electricity is alternating current with a frequency of 60 Hz. Its period is
B) amplitude
C) length
A) 60 cycles per second36. If you apply a force to an oscillator at its natural frequency, you will produce motion
B) 120 cycles per second
C) 0.0333 s
D) 0.0167 s; T = 1/f = 1/(60 Hz) = 1/(60 cyc/s) = (1/60) s = 0.0167 s
A) at exactly twice that frequency37. If a carefully calibrated pendulum were over a very large oil deposit, where the acceleration due to gravity is slightly decreased, what would happen to the pendulum's period?
B) at exactly one-half that frequency
C) with an amplitude that dies out or gets smaller.
D) with large amplitude
A) increase; T = 238. Like a transverse wave, a longitudinal wave has a/an
B) stay the same
C) decrease
A) amplitude39. For standing waves, antinodes
B) frequency
C) wavelength
D) all of the above
A) are half a wavelength apart40. On a string that is 1.0 m long, standing waves may be formed with the following wavelengths:
B) have the greatest amplitude
C) alternate with nodes
D) all of the above
A) 1.0 m, 2.0 m, 3.0 m
B) 1.0 m, 2.0 m, 4.0 m
C) 3.0 m, 1.5 m, 0.75 m
D) 2.0 m, 1.0 m, 0.5 m
(C) 1999 Doug Davis, all rights reserved