Questions Flashcards
Sound waves continue to travel until:
a. they are redirected by material surfaces.
b. they are completely dissipated by the effects of beam divergence.
c. they are transformed into another waveform.
d. all of the energy is converted into positive and negative ions.
a. they are redirected by material surfaces.
- Wavelength may be defined as:
a. frequency divided by velocity.
b. the distance along a wavetrain from peak to trough.
c. the distance from one point to the next identical point along the waveform.
d. the distance along a wavetrain from an area of high particle motion to one of low particle motion.
c. the distance from one point to the next identical point along the waveform.
- To determine wavelength:
a. multiply velocity by frequency.
b. divide velocity by frequency.
c. divide frequency by velocity.
d. multiply frequency by wavelength.
b. divide velocity by frequency.
- The wavelength of a 5 MHz sound wave in water is [VL = 1.483(10)5 cm/s]:
a. 0.297 mm (0.012 in.).
b. 2.54 mm (0.10 in.).
c. 296 mm (11.65 in.).
d. 3.00 mm (0.12 in.).
a. 0.297 mm (0.012 in.).
- Thickness resonance occurs when transducers and test parts are excited at a frequency equal to
(where V = sound velocity and T = item thickness):
a. 2T/V.
b. T/2V.
c. V/2T.
d. 2V/T.
c. V/2T.
- The equations that show VL and VT being dependent on elastic properties suggest that:
a. materials with higher densities will usually have higher acoustic velocities.
b. materials with higher moduli will usually have higher velocities.
c. wave velocities rely mostly upon the ratios of elastic moduli to material density.
d. VT will always be one-half of VL in the same material.
c. wave velocities rely mostly upon the ratios of elastic moduli to material density.
- Velocity measurements in a material revealed that the velocity decreased as frequency increased. This material
is called:
a. dissipated.
b. discontinuous.
c. dispersive.
d. degenerative.
c. dispersive.
- Plate thickness = 25.4 mm (1 in.), pulse-echo straight beam measured elapsed time = 8 µs. What is the most
likely material?
a. carbon steel.
b. lead.
c. titanium.
d. aluminum.
d. aluminum.
- The acoustic energy reflected at a plastic glass-quartz interface is equal to:
a. 64%.
b. 41%.
c. 22%.
d. 52%.
b. 41%.
- It can be deduced from Table 2 that the densities of:
a. plastic glass and water are in the ratio of 1.17:1.
b. steel and aluminum are in the ratio of 2.31:1.
c. quartz and aluminum are in the ratio of 1.05:1.
d. water and quartz are in the ratio of 10.13:1.
a. plastic glass and water are in the ratio of 1.17:1.
- The acoustic energy transmitted through a plastic glass-water interface is equal to:
a. 87%.
b. 36%.
c. 13%.
d. 64%
a. 87%.
- The first critical angle at a water-steel interface will be:
a. 18°.
b. 14.5°.
c. 22°.
d. 35°.
b. 14.5°.
- The second critical angle at a water-aluminum interface will be:
a. 28°.
b. 33°.
c. 67°.
d. 90°.
a. 28°.
- The incident angle needed in immersion testing to develop a 70° shear wave in plastic glass using the
information in Table 2 equals:
a. 83°.
b. 77°.
c. 74°.
d. 65°
b. 77°.
- Figure 2 shows the partition of incident and transmitted waves at a water-aluminum interface. At an
incident angle of 20°, the reflected wave and transmitted waves are respectively:
a. 60% and 40%.
b. 40% and 60%.
c. 1/3 and 2/3.
d. 80% and 20%.
a. 60% and 40%.
- From Figure 2 it is evident that the sum of the incident wave’s partitions (transmitted and reflected) is:
a. highly irregular at low angles, but constant above 30°.
b. lower at angles between 16° and 26°.
c. rarely more than 0.8.
d. always equal to unity.
d. always equal to unity.
- The principal attenuation modes are:
a. absorption, scatter, beam spread.
b. beam spread, collimation, scatter.
c. scatter, absorption, focusing.
d. scatter, beam spread, adhesion.
a. absorption, scatter, beam spread.
- Attenuation caused by scattering:
a. increases with increased frequency and grain size.
b. decreases with increased frequency and grain size.
c. increases with higher frequency and decreases with larger grain size.
d. decreases with higher frequency and decreases with larger grain size.
a. increases with increased frequency and grain size.
- In very fine-grain, isotropic crystalline material, the principal loss mechanism at 2 MHz is:
a. scatter.
b. mechanical hysteresis.
c. beam spread.
d. absorption.
c. beam spread.
- Two plates yield different backwall reflections in pulse-echo testing (18 dB) with their only apparent
difference being in the second material’s void content. The plates are both 75 mm (3 in.) thick. What is the
effective change in acoustic attenuation between the first and second plate based on actual metal path distance?
a. 0.118 dB/mm (3 dB/in.)
b. 0.236 dB/mm (6 dB/in.)
c. 0.709 dB/mm (18 dB/in.)
d. 0.039 dB/mm (1 dB/in.)
a. 0.118 dB/mm (3 dB/in.)
- The equation, sin ϕ = 0.7 λ/D, describes:
a. beam spread angle at 50% decrease in signal from the centerline value.
b. one-half the beam spread angle at 50% decrease in signal from the centerline value.
c. one-half the beam spread angle at 20% decrease in signal from the centerline value.
d. one-half the beam spread angle at 100% decrease in signal from the centerline value.
b. one-half the beam spread angle at 50% decrease in signal from the centerline value.
- The beam spread half-angle in the far field of a 25.4 mm (1 in.) diameter transducer sending 5 MHz longitudinal
waves into a plastic glass block is:
a. 0.5°.
b. 1.5°.
c. 3.1°.
d. 6.2°.
b. 1.5°.
- The near field of a round 12.7 mm (0.5 in.) diameter contact L-wave transducer being used on a steel test
part operating at 3 MHz is:
a. 12.7 mm (0.5 in.).
b. 25.4 mm (1 in.).
c. 9.9 mm (0.39 in.).
d. 20 mm (0.79 in.).
d. 20 mm (0.79 in.).
- The depth of penetration of the sound beam into a material can be increased by:
a. using a higher frequency.
b. using a longer wavelength.
c. using a smaller transducer.
d. using a lower frequency and a larger transducer.
d. using a lower frequency and a larger transducer.
- Barium titanate is a piezoelectric material which:
a. occurs naturally.
b. is piezoelectric at temperatures above the critical temperature.
c. has a high acoustic impedance.
d. is highly soluble in water.
c. has a high acoustic impedance.
- During an immersion test, numerous bubbles are noted in the water attached to the test item. These bubbles are small relative to the part size. What steps should the operator take?
a. Remove the bubbles by blowing them off with an air hose.
b. Ignore the bubbles because they are small and will not interfere with the test.
c. Remove the bubbles, with a brush or other mechanical means such as rubbing with the hand while the test is stopped.
d. Count the bubbles and mark their echoes on the test record.
c. Remove the bubbles, with a brush or other mechanical means such as rubbing with the hand while the test is stopped.
- A couplant is needed for a test on a hot steel plate (121 °C, 250 °F). Which of the following materials can be used?
a. water.
b. mercury.
c. tractor oil.
d. high-temperature grease.
d. high-temperature grease.
- A couplant is needed for a test on stainless steel welds. Numerous couplants are available. Which should be chosen and why?
a. a couplant free of chlorine because this element corrodes stainless steel.
b. glycerin because it is nonflammable.
c. oil because it is easily removed.
d. water because stainless steel does not corrode in water.
a. a couplant free of chlorine because this element corrodes stainless steel.
- A 5 MHz, 12.7 mm (0.5 in.) diameter, flat search unit in water has a near field length of approximately:
a. 177.8 mm (7 in.).
b. 50.8 mm (2 in.).
c. 84.6 mm (3-1/3 in.).
d. 139.7 mm (5-1/2 in.).
d. 139.7 mm (5-1/2 in.).
- A concave lens on a transducer will result in the near field in water being:
a. twice as long as with a flat lens.
b. three times as long as with a flat lens.
c. the same length as with a flat lens.
d. shorter than with a flat lens.
d. shorter than with a flat lens.
- A 10 MHz, 12.7 mm (0.5 in.) diameter search unit is placed on steel and acrylic plastic in succession. The
beam spread in these two materials is approximately:
a. 3° and 1.5°, respectively.
b. 1.5° and 3°, respectively.
c. equal in the two materials.
d. less than the beam spread of a 15 MHz search unit of the same diameter.
a. 3° and 1.5°, respectively.
- Focused transducers are useful because the:
a. smaller beam diameter increases the number of scans required to examine an object.
b. lateral resolution is improved.
c. lateral resolution is unimportant.
d. focal point is located beyond the end of the near field length of a similar, unfocused transducer.
b. lateral resolution is improved.
- Which of the following is a true statement about a sound beam with a longer wavelength.
a. A longer wavelength has better penetration than a shorter wavelength.
b. A longer wavelength provides a greater sensitivity and resolution.
c. A longer wavelength has less energy than a shorter wavelength.
d. Wavelength does not affect penetration, resolution or sensitivity.
a. A longer wavelength has better penetration than a shorter wavelength.
- Backing material on a transducer is used to:
a. damp the pulse and absorb the sound from the back of the transducer.
b. decrease the thickness oscillations.
c. increase the radial mode oscillations.
d. increase the power of the transmitted pulse.
a. damp the pulse and absorb the sound from the back of the transducer.
- Angle beam search units are used to:
a. inspect butt joint welds in thick-wall steel piping.
b. inspect pipe walls for internal corrosion.
c. examine material for acoustic velocity changes.
d. determine acoustic diffraction.
a. inspect butt joint welds in thick-wall steel piping.
- An angle beam transducer produces a 45° shear wave in steel. What is the approximate incident angle?
(velocity in steel = 0.125 in./µs, velocity in plastic = 0.105 in./µs; velocity in steel = 3.175 mm/µs,
velocity in plastic = 2.667 mm/µs)
a. 54.9°
b. 19°
c. 36.4°
d. 45°
c. 36.4°
- In Figure 6, the aluminum rod being examined is 152.4 mm (6 in.) in diameter. What is the offset
distance needed for a 45° refracted shear wave to be generated?
[L-wave velocity in aluminum = 6.3 (10)6 mm/s, T-wave velocity in aluminum = 3.1 (10)6 mm/s,
velocity in water = 1.5 (10)6 mm/s]
a. 5.13 mm (0.2 in.)
b. 26.06 mm (1.026 in.)
c. 52.12 mm (2.052 in.)
d. 15.05 mm (0.59 in.)
b. 26.06 mm (1.026 in.)
- In Figure 6 and using the conditions of question 13, what is the offset distance needed for a 45° refracted
longitudinal wave to be generated?
a. 10.03 mm (0.395 in.)
b. 4.5 mm (0.177 in.)
c. 12.82 mm (0.505 in.)
d. 10.26 mm (0.404 in.)
c. 12.82 mm (0.505 in.)
- It is desired to detect discontinuities 6.35 mm (0.25 in.) or less from the entry surface using angle beam shear waves. The search unit must be selected with the choice between a narrow band and a broadband unit. Which should be chosen and why?
a. The narrow band unit because it examines only a narrow band of the material.
b. The broadband unit because the entire volume is examined with a long pulse.
c. The broadband unit because the near surface resolution is better.
d. The broadband unit because the lateral resolution is excellent.
c. The broadband unit because the near surface resolution is better.