Unit 2.1 Multiple Choice Extended Questions  Waves
1. Once waves are no longer interacting with each other, the waves
A. trade energies
B. one wave gets consumed by the other
C. they double in size
D. the waves return to their original sizes
Answer D...
2. Constructive interference results in a wave whose amplitude is __________ any of the contributing waves.
A. larger than
B. smaller than
C. same size as
D. double
Answer A...
3. The principle of wave superposition DOES NOT apply to interactions between
A. two waves
B. more than two waves
C. coherent waves
D. inchoerent waves
Answer D...
4. According to the principle of superposition, displacement caused by combining two waves is equal to the
A. average amplitude of the contributing waves
B. amplitude of the larger wave
C. equilibrium position
D. sum of the amplitudes of contributing waves
Answer D...
5 . When two waves are in phase, ____________ interference will occur.
A. constructive
B. destructive
C. no
D. little
Answer A...
6. In transverse waves, the motion of the particles is, ____________ to the direction of propogation of the wave.
A. parallel
B. perpendicular
C. opposite
D. tangential
Answer B...
7. Particles move in a circle for the following type of wave:
A. surface
B. transverse
C. electromagnetic
D. longitudinal
Answer A...
8. Which type of wave DOES NOT requie a medium to propagate?
A. surface waves
B. transverse waves C. electromagnetic waves
D. longitudinal waves
Answer C...
9. In longitudinal waves, the motion of the particles is ____________ to the direction of propogation of the wave.
A. parallel
B. perpendicular
C. opposite
D. tangential
Answer A...
10. Which of the following is propagated by waves:
A. matter
B. energy
C. both matter and energy
D. none of the above
Answer B...
11. The horizontal distance between successive crests of a wave is
A. frequency
B. amplitude
C. period
D. wavelength
Answer D...
12. The highest point in a wave is
A. crest
B. trough
C. frequency
D. wavelength
Answer A...
13. The maximum displacement from the mediums equilibrium position is called
A. frequency
B. amplitude
C. period
D. wavelength
Answer B...
14. The time it takes two successive crests to pass the same place is is called
A. frequency
B. amplitude
C. period
D. wavelength
Answer C...
15. A ball moves with simple harmonic motion between points A and B. The magnitude of the acceleration (α) of the ball at point C is 4.00 ms^{2}. Calculate the acceleration of the ball at point D? (Recall that for harmonic motion α =  ω^{2}x)
A. +4.00 ms^{2}
B. 8.00 ms^{2}
C. 4.00 ms^{2}
D. +8.00 ms^{2}
Answer B...
At C
α = ω^{2}x
ω^{2} = ^{α}/_{x}
ω^{2} = (^{4.00}/_{0.200})
ω^{2} = 20 rads s^{1}
At D
α = (0.400) (20 rads s^{1})
α = 8.00 ms^{2}
16. Sound travels at 340 ms^{1} in air and 5000 ms^{1} in rocks. The sound of an earthquake emerges into the air. Its
A. frequency remains the same but the wavelength is shorter.
B. frequency is higher but the wavelength remains the same.
C. frequency is lower but the wavelength remains the same.
D. frequency remains the same but the wavelength is longer.
Answer A... Frequency is unchanged as the rate of vibration is the same. So as in this case speed and hence wavelenght decreases.
17. The upper graph shows the displacement y vs. the distance x for a wave is traveling with a speed v along the x axis in the positive direction. The lower graph shows the displacement y vs the time t. Calculate the speed of the wave from the information in the graphs.
A. 4.0 ms^{1}
B. 1.0 ms^{1}
C. 0.80 ms^{1}
D. 10.00 ms^{1}
Answer B..
V = ^{λ}/_{T}
From Upper graph λ = 4m (length of one complete wave)
Grom lower graph T = 4s (time take for one complete cycle)
V = ^{4m}/_{4s}
v = 1.0 ms^{1}
. .
18. The graph above shows two identical pulses traveling in opposite directions along a string, each with a velocity of 0.5 m/s. After 8.0 s, the string will look like
A.
B.
C.
D.
Answer B..
After 8.0s each pulse would have moved by 4.0m i.e. (0.5 m/s. X 8.0 s). Recall that the pulses will pass through each other without modification.
19. Two loudspeakers 5m apart emit the same tone of wavelength 0.5m. A listener stands at X, 12m from the right speaker and then moves to Y which is also 12m in front of the left speaker. She would hear
A. Maximum at X, Maximum at Y B. Maximum at X, Minimum at Y
C. Minimum at X, Minimum at Y
D. Minimum at X, Maximum at Y
Answer A..
X is 13m away from the left speaker and Y is also 13m from the right speaker (by Pythagoras Theorem).
Number of wavelengths between Y and left speaker = ^{12m}/_{0.5m} = 24
Number of wavelengths between Y and right speaker = ^{13m}/_{0.5m} = 26
Difference in wavelengths in both parts =2
Points that are nλ apart where n is an integer are in phase  Constructive interference (maximum at X)
A. Y is 13m away from the left speaker and Y is also 12m from the right speaker (by Pythagoras Theorem).
Number of wavelengths between Y and left speaker = ^{13m}/_{0.5m} = 26
Number of wavelengths between Y and right speaker = ^{12m}/_{0.5m} = 24
Difference in wavelengths in both parts =2
Points that are nλ apart where n is an integer are in phase  Constructive interference (maximum at Y)
20. Two progressive waves with equal velocities and wavelengths are moving in opposite directions in a string. The standing wave will have
A. Nodes at W and Y
B. Nodes at X and Z
C. Nodes at W and X
D. Nodes at Y and Z
Answer B..
To form a standing wave one end of the string is fixed and the reflected wave of that end interferes with the incident wave to create a node at the reflection point. The reflected wave is exactly out of phase with the incident wave at that point and both waves have zero displacement here, howeverone displacement is upward and the other downward. Nodes at also at ^{nλ}/_{2} where n is an integer and is measured from the fixed end. Anti nodes are at ^{(2n1)λ}/_{4 }where n is an integer and is measured from the fixed end.
Since both waves are moving at the same speed, they will meet at in the same way as the incident an reflected pulse. Note that in the wave moving left the particles are moving up with zero displacement and the wave moving left the particles are moving down with zero displacement. Thus nodes will be at ^{nλ}/_{2} where n is an integer and is measured from the fixed end.
Points which satisfy this condition are Z and X where n = 1 and 2 respectively. Antinodes will be at W and Y and the unlabelled red point, where n = 3, 2 and 1 respectively. I am sure you can now do the analysis for the green side.
