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Question 1 of 30Physics  Simple Harmonic Motion and Waves
1. Question
1. Which of the following is equivalent to the period of a system undergoing simple harmonic motion?Correct / You marked this questionThis is a conceptual question.
Period means the amount of time it takes to complete one cycle.
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating spring is also equivalent to one wavelength.
However, for an oscillating spring, the “crest” of the wave would equal the maximum positive displacement of the spring, the farthest the spring is stretched (or compressed). The “trough” of the wave would equal the maximum negative displacement of the spring, the farthest the spring travels in the opposite direction once it is released.
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.
Incorrect / You marked this questionThis is a conceptual question.
Period means the amount of time it takes to complete one cycle.
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating spring is also equivalent to one wavelength.
However, for an oscillating spring, the “crest” of the wave would equal the maximum positive displacement of the spring, the farthest the spring is stretched (or compressed). The “trough” of the wave would equal the maximum negative displacement of the spring, the farthest the spring travels in the opposite direction once it is released.
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.

Question 2 of 30Physics  Simple Harmonic Motion and Waves
2. Question
2. In terms of simple harmonic motion, period is in units of:Correct / You marked this questionThis is a conceptual question.
Period means the amount of time it takes to complete one cycle. Period could be measured in any units of time (seconds, minutes, hours, days, etc.).
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating spring is also equivalent to one wavelength.
However, for an oscillating spring, the “crest” of the wave would equal the maximum positive displacement of the spring, the farthest the spring is stretched (or compressed). The “trough” of the wave would equal the maximum negative displacement of the spring, the farthest the spring travels in the opposite direction once it is released.
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.
Incorrect / You marked this questionThis is a conceptual question.
Period means the amount of time it takes to complete one cycle. Period could be measured in any units of time (seconds, minutes, hours, days, etc.).
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating spring is also equivalent to one wavelength.
However, for an oscillating spring, the “crest” of the wave would equal the maximum positive displacement of the spring, the farthest the spring is stretched (or compressed). The “trough” of the wave would equal the maximum negative displacement of the spring, the farthest the spring travels in the opposite direction once it is released.
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.

Question 3 of 30Physics  Simple Harmonic Motion and Waves
3. Question
3. Which of the following depends on the mass of an attached object?Correct / You marked this questionLet’s look at each of our answer options individually. For each, determining dependence on mass will mean recalling the respective equations:
[A] The spring constant, k
[B] The angular frequency of an oscillating pendulum
The angular frequency of a pendulum inversely proportional to the period of the pendulum. The period of a pendulum is NOT dependent on mass, thus, neither is the angular frequency of a pendulum.
[C] The period of an oscillating pendulum
[D] The period of an oscillating spring
This is the correct answer. You can see that the period of an oscillating spring is dependent on mass in its equation:
[E] The potential energy of an oscillating spring
Incorrect / You marked this questionLet’s look at each of our answer options individually. For each, determining dependence on mass will mean recalling the respective equations:
[A] The spring constant, k
[B] The angular frequency of an oscillating pendulum
The angular frequency of a pendulum inversely proportional to the period of the pendulum. The period of a pendulum is NOT dependent on mass, thus, neither is the angular frequency of a pendulum.
[C] The period of an oscillating pendulum
[D] The period of an oscillating spring
This is the correct answer. You can see that the period of an oscillating spring is dependent on mass in its equation:
[E] The potential energy of an oscillating spring

Question 4 of 30Physics  Simple Harmonic Motion and Waves
4. Question
4. At what point is maximum kinetic energy achieved for an oscillating massspring system; where x is displacement from equilibrium?Correct / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum KE and PE points for a massspring, but it simpler/ more effective to just reason through intuitively.
[A] x = 1
x = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a spring could be displaced almost any possible value of length..
[B] x = maximum displacement
The maximum displacement for a spring would be at its most stretched or compressed. At these points, the object is not moving. Thus, KE = 0
[C] x = 0
x = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum.
[D] 0 < x < maximum displacement
When x is less than maximum displacement, it is either gaining kinetic energy (speed is increasing if the object is moving toward equilibrium) or losing kinetic energy (object is slowing down if moving away from equilibrium). In neither case has it achieved maximum KE.
[E] x > maximum displacement
x can never be greater than maximum displacement!
Incorrect / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum KE and PE points for a massspring, but it simpler/ more effective to just reason through intuitively.
[A] x = 1
x = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a spring could be displaced almost any possible value of length..
[B] x = maximum displacement
The maximum displacement for a spring would be at its most stretched or compressed. At these points, the object is not moving. Thus, KE = 0
[C] x = 0
x = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum.
[D] 0 < x < maximum displacement
When x is less than maximum displacement, it is either gaining kinetic energy (speed is increasing if the object is moving toward equilibrium) or losing kinetic energy (object is slowing down if moving away from equilibrium). In neither case has it achieved maximum KE.
[E] x > maximum displacement
x can never be greater than maximum displacement!

Question 5 of 30Physics  Simple Harmonic Motion and Waves
5. Question
5. At what point is maximum potential energy achieved for an oscillating massspring system; where x is displacement from equilibrium?Correct / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum PE and KE points for a massspring, but it simpler/ more effective to just reason through intuitively.
[A] x = 1
x = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a spring could be displaced almost any possible value of length.
[B] x = maximum displacement
The maximum displacement for a spring wouldbe at its most stretched or compressed. At these points, the object is not moving. Thus, KE = 0, and PE is at its maximum.
[C] x = 0
x = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum. And PE is at its minimum
[D] 0 < x < maximum displacement
When x is less than maximum displacement, it is either losing PE (if the object is moving toward equilibrium) or gaining PE (if the object is moving away from equilibrium). In neither case has it achieved maximum PE.
[E] x > maximum displacement
x can never be greater than maximum displacement!
Incorrect / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum PE and KE points for a massspring, but it simpler/ more effective to just reason through intuitively.
[A] x = 1
x = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a spring could be displaced almost any possible value of length.
[B] x = maximum displacement
The maximum displacement for a spring wouldbe at its most stretched or compressed. At these points, the object is not moving. Thus, KE = 0, and PE is at its maximum.
[C] x = 0
x = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum. And PE is at its minimum
[D] 0 < x < maximum displacement
When x is less than maximum displacement, it is either losing PE (if the object is moving toward equilibrium) or gaining PE (if the object is moving away from equilibrium). In neither case has it achieved maximum PE.
[E] x > maximum displacement
x can never be greater than maximum displacement!

Question 6 of 30Physics  Simple Harmonic Motion and Waves
6. Question
6. At what point is maximum kinetic energy achieved for an oscillating pendulum system? Let θ be the displacement angle from equilibrium.Correct / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum KE and PE points a pendulum, but it simpler/ more effective to just reason through intuitively.
[A] θ = 1
θ = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a pendulum could be displaced almost any possible angular value.
[B] θ = θ_{MAX}
θ_{MAX}, the maximum displacement for a pendulum, is when the pendulum has traveled farthest to one side or the other. At these points, the object is not moving. Thus, KE = 0
[C] θ = 0
θ = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum.
[D] 0 < θ < θ_{MAX}
When θ is less than θ_{MAX}, it is either gaining kinetic energy (speed is increasing if the object is moving toward equilibrium) or losing kinetic energy (object is slowing down if moving away from equilibrium). In neither case has it achieved maximum KE.
[E] θ > θ_{MAX}
θ can never be greater than θ_{MAX}!
Incorrect / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum KE and PE points a pendulum, but it simpler/ more effective to just reason through intuitively.
[A] θ = 1
θ = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a pendulum could be displaced almost any possible angular value.
[B] θ = θ_{MAX}
θ_{MAX}, the maximum displacement for a pendulum, is when the pendulum has traveled farthest to one side or the other. At these points, the object is not moving. Thus, KE = 0
[C] θ = 0
θ = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, KE is at its maximum.
[D] 0 < θ < θ_{MAX}
When θ is less than θ_{MAX}, it is either gaining kinetic energy (speed is increasing if the object is moving toward equilibrium) or losing kinetic energy (object is slowing down if moving away from equilibrium). In neither case has it achieved maximum KE.
[E] θ > θ_{MAX}
θ can never be greater than θ_{MAX}!

Question 7 of 30Physics  Simple Harmonic Motion and Waves
7. Question
7. At what point is maximum potential energy achieved for an oscillating pendulum system? Let θ be the displacement angle from equilibrium.Correct / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum PE and KE points a pendulum, but it simpler/ more effective to just reason through intuitively.
[A] θ = 1
θ = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a pendulum could be displaced almost any possible angular value.
[B] θ = θ_{MAX}
θ_{MAX}, the maximum displacement for a pendulum, is when the pendulum has traveled farthest to one side or the other. At these points, the object is not moving. Thus, KE = 0, and potential energy is at its maximum.
[C] θ = 0
θ = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, PE is at its minimum.
[D] 0 < θ < θ_{MAX}
When θ is less than θ_{MAX}, it is either losing PE (if the object is moving toward equilibrium) or gaining PE (object is slowing down if moving away from equilibrium). In neither case has it reached maximum PE.
[E] θ > θ_{MAX}
θ can never be greater than θ_{MAX}!
Incorrect / You marked this questionLet’s look at each of our answer options individually. Technically, you could memorize the maximum PE and KE points a pendulum, but it simpler/ more effective to just reason through intuitively.
[A] θ = 1
θ = 1 is an arbitrary displacement value. You may think of 1 as the maximum value for the sine wave. However, a pendulum could be displaced almost any possible angular value.
[B] θ = θ_{MAX}
θ_{MAX}, the maximum displacement for a pendulum, is when the pendulum has traveled farthest to one side or the other. At these points, the object is not moving. Thus, KE = 0, and potential energy is at its maximum.
[C] θ = 0
θ = 0 is the equilibrium position. At this point, all of the system’s potential energy has been converted into kinetic energy. Thus, PE is at its minimum.
[D] 0 < θ < θ_{MAX}
When θ is less than θ_{MAX}, it is either losing PE (if the object is moving toward equilibrium) or gaining PE (object is slowing down if moving away from equilibrium). In neither case has it reached maximum PE.
[E] θ > θ_{MAX}
θ can never be greater than θ_{MAX}!

Question 8 of 30Physics  Simple Harmonic Motion and Waves
8. Question
8. Jeremy buys a spring from the hardware store. The spring is 3 meters long. In order to stretch the spring 1.2 m, Jeremy must apply 60 N of force. How much force must he apply to stretch the spring 0.7 m?Correct / You marked this questionStep 1: Let’s recall our SpringForce equation:
Step 2: Solve for k
The problem gives you F & x in the first situation. We can use these to solve for k, which we can then use to solve for F in the second situation. The important concept here is the k is constant for the spring, regardless of how much or how forcefully it is stretched/ compressed.
F_{1} = kx
60 N = k (1.2 m)
k = 50 N/mStep 2: Solve for F
F_{2} = kx
F_{2} = (50 N/m)(0.7 m)
F_{2} = 35 NIncorrect / You marked this questionStep 1: Let’s recall our SpringForce equation:
Step 2: Solve for k
The problem gives you F & x in the first situation. We can use these to solve for k, which we can then use to solve for F in the second situation. The important concept here is the k is constant for the spring, regardless of how much or how forcefully it is stretched/ compressed.
F_{1} = kx
60 N = k (1.2 m)
k = 50 N/mStep 2: Solve for F
F_{2} = kx
F_{2} = (50 N/m)(0.7 m)
F_{2} = 35 N 
Question 9 of 30Physics  Simple Harmonic Motion and Waves
9. Question
9. Energy from Planet X is emitted through its inner atmosphere. The energy has wavelength 4,000 nm and frequency of
4 x 10^{4} Hz.
What is the speed of this energy?Correct / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wave velocity
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
v = (4 x 10^{4} wavelengths/second)(4,000 nm/wavelength)
v = (4 x 10^{4} wavelengths/second)(4 x 10^{6} m/wavelength)
v = 0.16 m/sIncorrect / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wave velocity
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
v = (4 x 10^{4} wavelengths/second)(4,000 nm/wavelength)
v = (4 x 10^{4} wavelengths/second)(4 x 10^{6} m/wavelength)
v = 0.16 m/s 
Question 10 of 30Physics  Simple Harmonic Motion and Waves
10. Question
10. A wave has velocity 3.5 m/s and frequency of 5 x 10^{6} Hz. What is the wavelength of the wave?Correct / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wavelength
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3.5 m/s = (5 x 10^{6} wavelengths/second) λ
λ = 7 x 10^{7} m
λ = 700 nmRemember: 1 m = 1 x 10^{9} nm
Incorrect / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wavelength
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3.5 m/s = (5 x 10^{6} wavelengths/second) λ
λ = 7 x 10^{7} m
λ = 700 nmRemember: 1 m = 1 x 10^{9} nm

Question 11 of 30Physics  Simple Harmonic Motion and Waves
11. Question
11. UV rays have a wavelength 20 nm of and frequency of 3 x 10^{14} Hz. What is the speed of these UV rays?Correct / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wavelength
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
v = (3 x 10^{14} wavelengths/second)(20 nm)
v = (3 x 10^{14} wavelengths/second)(2 x 10^{8} m)
v = 6 x 10^{6} m/sRemember: 1 nm = 1 x 10^{9} m
Incorrect / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for wavelength
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
v = (3 x 10^{14} wavelengths/second)(20 nm)
v = (3 x 10^{14} wavelengths/second)(2 x 10^{8} m)
v = 6 x 10^{6} m/sRemember: 1 nm = 1 x 10^{9} m

Question 12 of 30Physics  Simple Harmonic Motion and Waves
12. Question
12. Microwaves have a speed of 3 x 10^{8} m/s and wavelength of 1 m. What is the frequency of these microwaves?Correct / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for frequency
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3 x 10^{8} m/s = (1 m/wavelength)(wavelengths/second)
f = 3 x 10^{8} wavelengths/second = 3 x 10^{8} Hz
3 x 10^{8} Hz = 300 MHzRemember: 1 Mhz = 1 x 10^{6} Hz
Incorrect / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for frequency
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3 x 10^{8} m/s = (1 m/wavelength)(wavelengths/second)
f = 3 x 10^{8} wavelengths/second = 3 x 10^{8} Hz
3 x 10^{8} Hz = 300 MHzRemember: 1 Mhz = 1 x 10^{6} Hz

Question 13 of 30Physics  Simple Harmonic Motion and Waves
13. Question
13. What is the wavelength of the following wave?
Correct / You marked this questionStep 1: Recall what exactly is a wavelength
One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
Step 2: Determine the length of one wavelength
We can see that 10 m equates to the distance between one crest (at 0) and another crest (at 10 m). Alternatively, we could count that distance from 0 to 10 m includes 1 hump below the x axis, and 2 “half humps” above the x axis. I.e. one total up hump, and one down hump.
Thus, wavelength = 10 m
Incorrect / You marked this questionStep 1: Recall what exactly is a wavelength
One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
Step 2: Determine the length of one wavelength
We can see that 10 m equates to the distance between one crest (at 0) and another crest (at 10 m). Alternatively, we could count that distance from 0 to 10 m includes 1 hump below the x axis, and 2 “half humps” above the x axis. I.e. one total up hump, and one down hump.
Thus, wavelength = 10 m

Question 14 of 30Physics  Simple Harmonic Motion and Waves
14. Question
14. Radio waves have a speed of 3 x 10^{6} m/s and wavelength of 1 km. What is the period of these radio waves?Correct / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for frequency
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3 x 10^{6} m/s = (1000 m/wavelength)(wavelengths/second)
f = (3 x 10^{6} m/s) / (1000 m/wavelength)
f = 3,000 wavelengths/secondStep 3: Solve for period
Recall our period equation:
T = 1 / f
T = ^{1}/_{3,000} secondsIncorrect / You marked this questionStep 1: Let’s recall our wave velocity equation:
Step 2: Solve for frequency
Remember: Hz are equivalent to cycles per second, i.e. wavelengths/second. We’ll write wavelengths/second in place of Hz to make the calculation easier
v = λf
3 x 10^{6} m/s = (1000 m/wavelength)(wavelengths/second)
f = (3 x 10^{6} m/s) / (1000 m/wavelength)
f = 3,000 wavelengths/secondStep 3: Solve for period
Recall our period equation:
T = 1 / f
T = ^{1}/_{3,000} seconds 
Question 15 of 30Physics  Simple Harmonic Motion and Waves
15. Question
15. You have a spring that is 2.28 meters long. In order to compress the spring 0.74 m, you must apply 40.5 N of force. What is the relative work required to compress the spring 2x as much?Correct / You marked this questionStep 1: Let’s recall our SpringWork equation:
You can see that as long as we hold the other factors constant (i.e. we are using the same spring) it will require (2x)^{2} the work if we double the distance compressed.
That is, the relative work required will be 4x.
Incorrect / You marked this questionStep 1: Let’s recall our SpringWork equation:
You can see that as long as we hold the other factors constant (i.e. we are using the same spring) it will require (2x)^{2} the work if we double the distance compressed.
That is, the relative work required will be 4x.

Question 16 of 30Physics  Simple Harmonic Motion and Waves
16. Question
16. You have a spring that is 2.67 meters long. k = 11 N/m. In order to compress the spring 0.74 m, you must apply 40.5 N of force. If you held all else the same, but doubled the spring length, how much relative work would be required to accomplish the same task?Correct / You marked this questionStep 1: Let’s recall our SpringWork equation:
You can see that the work required to stretch or compressed a spring is NOT dependent on spring length.
Thus, the relative work required will be the same.
Incorrect / You marked this questionStep 1: Let’s recall our SpringWork equation:
You can see that the work required to stretch or compressed a spring is NOT dependent on spring length.
Thus, the relative work required will be the same.

Question 17 of 30Physics  Simple Harmonic Motion and Waves
17. Question
17. You have a spring that is 40 cm. k = 5 N/m. You compress the spring 20 cm. What is the potential energy of the spring?Correct / You marked this questionStep 1: Let’s recall our SpringPotential Energy equation:
Step 2: Solve for PE
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(5 N/m)(0.2 m)^{2}
PE = 0.1 JIncorrect / You marked this questionStep 1: Let’s recall our SpringPotential Energy equation:
Step 2: Solve for PE
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(5 N/m)(0.2 m)^{2}
PE = 0.1 J 
Question 18 of 30Physics  Simple Harmonic Motion and Waves
18. Question
18. A 160 m pendulum, with a ball of mass 16 kg attached, is swaying back and forth. The pendulum reached a maximum angle of displacement (θ_{MAX}) of 45^{o} from equilibrium. What is the period of the pendulum?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s recall our PeriodPendulum Equation
Step 2: Solve for period
Step 3: Let’s recall what period means for an oscillating pendulum.
Period means the amount of time it takes to complete one cycle.
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.
Incorrect / You marked this questionStep 1: Let’s recall our PeriodPendulum Equation
Step 2: Solve for period
Step 3: Let’s recall what period means for an oscillating pendulum.
Period means the amount of time it takes to complete one cycle.
One cycle of a wave is equivalent to one wavelength. One wavelength equates to the distance between crest to crest, trough to trough, or two “humps.”
One cycle of an oscillating pendulum is also equivalent to one wavelength.
However, for an oscillating pendulum, the “crest” of the wave would equal the maximum positive displacement of the pendulum, the farthest the pendulum travels to one side. The “trough” of the wave would equal the maximum negative displacement of the pendulum, the farthest the pendulum travels to the opposite side.

Question 19 of 30Physics  Simple Harmonic Motion and Waves
19. Question
19. You have a spring of 50 cm that is attached to a mass. k = 5 N/m. You compress the spring 10 cm. What is the period of the spring?Correct / You marked this questionLet’s recall our SpringPeriod Energy equation:
As you can see: the period of a springmass system is DEPENDENT on the attached mass! We are not given the mass of the attached object. Thus, we cannot determine the the period of the spring.
Incorrect / You marked this questionLet’s recall our SpringPeriod Energy equation:
As you can see: the period of a springmass system is DEPENDENT on the attached mass! We are not given the mass of the attached object. Thus, we cannot determine the the period of the spring.

Question 20 of 30Physics  Simple Harmonic Motion and Waves
20. Question
20. You have a spring of 80 cm that is attached to a mass of 9 kg. k = 1 N/m. You compress the spring 20 cm. What is the period of the spring?Correct / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
Step 2: Solve for Period
Incorrect / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
Step 2: Solve for Period

Question 21 of 30Physics  Simple Harmonic Motion and Waves
21. Question
21. Which of the following is true about simple harmonic motion?Correct / You marked this questionThis is a conceptual question. Let’s analyze each answer option individually:
[A] Restoring force acts in the direction of displacement
You can reason this one out intuitively. When you stretch / compress a spring in one direction, its restoring force pulls it in the opposite direction. Same idea for a pendulum. Thus, restoring force acts in the OPPOSITE direction of displacement.
[B] Restoring force is proportional to the equilibrium force
This is a nonsense answer. Equilibrium force has no meaning in this context.
[C] Restoring force is proportional to the mass of the displaced object
Recall our SpringForce equation (i.e Hooke’s Law):
You can see that the restoring force (F) is not dependent on the mass of the displaced object.
[D] Restoring force is proportional to displacement from equilibrium
This is the correct answer. Recall our SpringForce equation (i.e Hooke’s Law):
You can see the restoring force (F) is directly proportional to the displacement from equilibrium (x).
[E] None of the above
D is the correct answer
Incorrect / You marked this questionThis is a conceptual question. Let’s analyze each answer option individually:
[A] Restoring force acts in the direction of displacement
You can reason this one out intuitively. When you stretch / compress a spring in one direction, its restoring force pulls it in the opposite direction. Same idea for a pendulum. Thus, restoring force acts in the OPPOSITE direction of displacement.
[B] Restoring force is proportional to the equilibrium force
This is a nonsense answer. Equilibrium force has no meaning in this context.
[C] Restoring force is proportional to the mass of the displaced object
Recall our SpringForce equation (i.e Hooke’s Law):
You can see that the restoring force (F) is not dependent on the mass of the displaced object.
[D] Restoring force is proportional to displacement from equilibrium
This is the correct answer. Recall our SpringForce equation (i.e Hooke’s Law):
You can see the restoring force (F) is directly proportional to the displacement from equilibrium (x).
[E] None of the above
D is the correct answer

Question 22 of 30Physics  Simple Harmonic Motion and Waves
22. Question
22. You have a spring of 100 cm that is attached to a mass of 50 kg. k = 0.5 N/m. You keep all else the same, but halve the spring length, and double the mass of the object, what is the relative period of the spring?Correct / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
You can see that the period of a spring is NOT dependent on spring length, but it IS dependent on the mass of the attached object.
Step 2: Solve for Period in original and altered scenario
Original
Altered
Thus, you can see the relative period is √2x
Incorrect / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
You can see that the period of a spring is NOT dependent on spring length, but it IS dependent on the mass of the attached object.
Step 2: Solve for Period in original and altered scenario
Original
Altered
Thus, you can see the relative period is √2x

Question 23 of 30Physics  Simple Harmonic Motion and Waves
23. Question
23. You have a spring of 4 m that is attached to a mass of 10 kg. k = 0.4 N/m. You compress the spring 2 m in 0.4 seconds. What is the power required to cause the compression?Correct / You marked this questionStep 1: Let’s recall our SpringWork Equation:
Our equation sheet does not specifically list a SpringPower equation, but we know that power is simply work / time. We also know that power is typically in units of watts.
Step 2: Solve for Power
Incorrect / You marked this questionStep 1: Let’s recall our SpringWork Equation:
Our equation sheet does not specifically list a SpringPower equation, but we know that power is simply work / time. We also know that power is typically in units of watts.
Step 2: Solve for Power

Question 24 of 30Physics  Simple Harmonic Motion and Waves
24. Question
24. A spring is hanging from the underside of a table. The 300 cm spring stretches 100 cm when a block of mass 2 kg is attached. What is the period if everything else is held the same, but the block is 80 kg?Correct / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
We are given the mass of the spring, but not k. Thus, we must solve for k in order to solve for period.
Step 2: Solve for k
It might not be immediately obvious, but this force (F) will be equal in magnitude to the weight of the block.
F = kx = F_{w} = mg
kx = mg
kx = (2 kg)(10 m/s^{2})
k = (2 kg)(10 m/s^{2}) / x
k = (2 kg)(10 m/s^{2}) / (1 m)
k = 20 N/mStep 3: Solve for period with 8 kg block
Incorrect / You marked this questionStep 1: Let’s recall our SpringPeriod Energy equation:
We are given the mass of the spring, but not k. Thus, we must solve for k in order to solve for period.
Step 2: Solve for k
It might not be immediately obvious, but this force (F) will be equal in magnitude to the weight of the block.
F = kx = F_{w} = mg
kx = mg
kx = (2 kg)(10 m/s^{2})
k = (2 kg)(10 m/s^{2}) / x
k = (2 kg)(10 m/s^{2}) / (1 m)
k = 20 N/mStep 3: Solve for period with 8 kg block

Question 25 of 30Physics  Simple Harmonic Motion and Waves
25. Question
25. An oscillating pendulum takes 40 seconds to complete 8 cycles back and forth. The pendulum has an attached mass of 20 kg, and a maximum displacement of θ_{MAX} = 60^{o}.
What is the period of the pendulum?
Assume gravitational acceleration (g) = 10 m/s^{2}Correct / You marked this questionStep 1: Let’s recall our PeriodPendulum equation:
You can quickly see that we are not given the length of the pendulum, and thus, cannot solve for the period of the pendulum this way.
Step 2: Let’s recall what period means for an oscillating pendulum.
Period means the amount of time it takes to complete one cycle. If it takes 40 seconds to complete 8 cycles, then it takes takes
(40 seconds) / (8 cycles) to complete 1 cycle, which equals 5 seconds/cycle.Incorrect / You marked this questionStep 1: Let’s recall our PeriodPendulum equation:
You can quickly see that we are not given the length of the pendulum, and thus, cannot solve for the period of the pendulum this way.
Step 2: Let’s recall what period means for an oscillating pendulum.
Period means the amount of time it takes to complete one cycle. If it takes 40 seconds to complete 8 cycles, then it takes takes
(40 seconds) / (8 cycles) to complete 1 cycle, which equals 5 seconds/cycle. 
Question 26 of 30Physics  Simple Harmonic Motion and Waves
26. Question
26. A mass is attached to a spring on a frictionless table. The spring is compressed 8 m. The mass of the attached object is 30 kg, and the spring constant k is 10 N/m.
What is the magnitude of the velocity of the object after the spring has released 4 m towards equilibrium?Correct / You marked this questionStep 1: Let’s recall our SpringPotential Energy equation:
Step 2: Solve for PE of the fully compressed spring
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(10 N/m)(8 m)^{2}
PE_{1} = 320 JNote: when the spring is fully compressed, KE = 0, and PE = Total Energy
Step 2: Solve for PE of the spring after releasing 4 m towards equilibrium
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(10 N/m)(4 m)^{2}
PE_{2} = 80 JStep 3: Determine how much PE has been converted to KE
Total Energy – PE_{2} = PE converted to KE
KE = 320 J – 80 J = 240 JStep 4: Determine velocity based on KE
Recall our KE equation:
KE = ^{1}/_{2}mv^{2}
240 J = ^{1}/_{2}(30 kg) v^{2}
v^{2} = 16 m^{2}s^{2}
v = 4 m/sIncorrect / You marked this questionStep 1: Let’s recall our SpringPotential Energy equation:
Step 2: Solve for PE of the fully compressed spring
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(10 N/m)(8 m)^{2}
PE_{1} = 320 JNote: when the spring is fully compressed, KE = 0, and PE = Total Energy
Step 2: Solve for PE of the spring after releasing 4 m towards equilibrium
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(10 N/m)(4 m)^{2}
PE_{2} = 80 JStep 3: Determine how much PE has been converted to KE
Total Energy – PE_{2} = PE converted to KE
KE = 320 J – 80 J = 240 JStep 4: Determine velocity based on KE
Recall our KE equation:
KE = ^{1}/_{2}mv^{2}
240 J = ^{1}/_{2}(30 kg) v^{2}
v^{2} = 16 m^{2}s^{2}
v = 4 m/s 
Question 27 of 30Physics  Simple Harmonic Motion and Waves
27. Question
27. Using a force of 10 N, a spring is stretched 10 cm. After it is released, and travels 6 cm towards equilibrium, its speed is
5 m/s.
What is the total energy of the system at this moment?
Discount friction and air resistance.Correct / You marked this questionStep 1: Let’s recall what total energy means
As long as we can discount friction and air resistance, we know that the total energy of a system DOESN’T CHANGE. PE may be converted to KE, or vice versa, but the total energy remains the same.
Step 2: Solve for PE of stretched spring
To solve for PE, we must solve for k
Step 3: Solve for k
F = kx
10 N = k (0.1m)
k = 100 N/mStep 3: Solve for PE
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(100 N/m)(0.1 m)^{2}
PE = Total Energy = 0.5 JIncorrect / You marked this questionStep 1: Let’s recall what total energy means
As long as we can discount friction and air resistance, we know that the total energy of a system DOESN’T CHANGE. PE may be converted to KE, or vice versa, but the total energy remains the same.
Step 2: Solve for PE of stretched spring
To solve for PE, we must solve for k
Step 3: Solve for k
F = kx
10 N = k (0.1m)
k = 100 N/mStep 3: Solve for PE
PE = ^{1}/_{2}kx^{2}
PE = ^{1}/_{2}(100 N/m)(0.1 m)^{2}
PE = Total Energy = 0.5 J 
Question 28 of 30Physics  Simple Harmonic Motion and Waves
28. Question
28. You turn up your stereo and increase the sound 10x in intensity. This would equate to an increase of:Correct / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity.
For example, for something that emits a sound of 20 dB:
Incorrect / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity.
For example, for something that emits a sound of 20 dB:

Question 29 of 30Physics  Simple Harmonic Motion and Waves
29. Question
29. You turn up your stereo and increase the sound 100x in intensity. This would equate to an increase of:Correct / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity. Thus, a 100 fold increase would equal
10 dB + 10 dB = 20 dB increase.For example, for something that emits a sound of 20 dB:
Incorrect / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity. Thus, a 100 fold increase would equal
10 dB + 10 dB = 20 dB increase.For example, for something that emits a sound of 20 dB:

Question 30 of 30Physics  Simple Harmonic Motion and Waves
30. Question
30. You turn up your stereo and increase the sound 30 decibels This would equate sound intensity increase of:Correct / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity. Thus, 30 fold increase would equal
10^{3} dB = 1000 dB increaseFor example, for something that emits a sound of 40 dB:
Incorrect / You marked this questionEvery increase of 10 decibels (dB) equates to a 10 fold increase in sound intensity. Thus, 30 fold increase would equal
10^{3} dB = 1000 dB increaseFor example, for something that emits a sound of 40 dB:
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