11th Grade Physics: Quiz On Simple Harmonic Motion!
Below is an 11th-grade physics trivia quiz on simple harmonic motion! One of the major rules when it comes to forcing is that for every action, there is a reaction. The simple harmonic motion is a special type of periodic motion where the restoring force on the moving object is directly proportional to and opposite the object's displacement vector. Do you know how to calculate force? Take this quiz and get to refresh your understanding. Do not forget to share the quiz with others, also.
Questions and Answers
- 1.
Which statement best describes simple harmonic motion?
- A.
An oscillation that increases in amplitude over time
- B.
An oscillation that gets quicker and quicker
- C.
An oscillation with a fixed period and fixed amplitude
- D.
An oscillation with fixed period and amplitude which varies over time
Correct Answer
C. An oscillation with a fixed period and fixed amplitudeExplanation
Simple harmonic motion is characterized by an oscillation with a fixed period and fixed amplitude. This means that the motion repeats itself in a regular pattern, with the same amount of time taken for each complete cycle and the same maximum displacement from the equilibrium position. The other options, such as increasing amplitude over time or getting quicker and quicker, do not accurately describe simple harmonic motion.Rate this question:
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- 2.
A particle undergoes simple harmonic motion with an angular velocity of 3 rad/s and amplitude of 30 cm. It starts with maximum forward amplitude at time t = 0. Find the velocity at time t = 2s.
- A.
0.251 m/s
- B.
3.21 m/s
- C.
0.545 m/s
- D.
0.114 m/s
Correct Answer
A. 0.251 m/sExplanation
The velocity of a particle undergoing simple harmonic motion can be calculated using the formula v = ωAcos(ωt + φ), where ω is the angular velocity, A is the amplitude, t is the time, and φ is the phase constant. In this case, the angular velocity is given as 3 rad/s and the amplitude is 30 cm. Since the particle starts with maximum forward amplitude at t = 0, the phase constant φ is 0. Plugging in these values and t = 2s into the formula, we get v = 3 * 0.3 * cos(3 * 2 + 0) = 0.251 m/s. Therefore, the velocity at t = 2s is 0.251 m/s.Rate this question:
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- 3.
A particle undergoes simple harmonic motion with an angular velocity of 5 rad/s and an amplitude of 50 cm. It starts with maximum forward amplitude at time t = 0. Find the displacement at time t = 10 s.
- A.
0.500 m
- B.
0.482 m
- C.
0.321 m
- D.
0.438 m
Correct Answer
B. 0.482 mExplanation
The given problem describes a particle undergoing simple harmonic motion with an angular velocity of 5 rad/s and an amplitude of 50 cm. The particle starts with maximum forward amplitude at time t = 0. To find the displacement at time t = 10 s, we can use the equation for displacement in simple harmonic motion:
x = A * cos(ωt)
Where x is the displacement, A is the amplitude, ω is the angular velocity, and t is the time. Plugging in the given values, we have:
x = 50 cm * cos(5 rad/s * 10 s)
Calculating this expression, we find that x ≈ 0.482 m. Therefore, the displacement at t = 10 s is approximately 0.482 m.Rate this question:
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- 4.
A particle undergoes simple harmonic motion with an angular velocity of 7 rad/s and amplitude of 2 m. It starts with maximum forward amplitude at time t = 0. Find the acceleration at time t = 6s.
- A.
39.2 m/s/s
- B.
1.23 m/s/s
- C.
82.4 m/s/s
- D.
9.81 m/s/s
Correct Answer
A. 39.2 m/s/sExplanation
The acceleration of a particle undergoing simple harmonic motion can be calculated using the formula a = -ω^2x, where ω is the angular velocity and x is the displacement from the equilibrium position. In this case, the angular velocity is 7 rad/s and the displacement is 2 m. Plugging these values into the formula, we get a = -(7^2)(2) = -98 m/s^2. Since the particle starts with maximum forward amplitude, the acceleration at time t = 6s will have the same magnitude but opposite direction. Therefore, the acceleration at t = 6s is 98 m/s^2, which is equivalent to 39.2 m/s/s.Rate this question:
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- 5.
Below is the expression for:
- A.
Acceleration
- B.
Velocity
- C.
Displacement
- D.
None of the above
Correct Answer
D. None of the above -
- 6.
Below is the expression for:
- A.
Acceleration
- B.
Velocity
- C.
Displacement
- D.
None of the above
Correct Answer
A. AccelerationExplanation
Acceleration is the correct answer because the question is asking for the expression related to acceleration, velocity, displacement, or none of the above. Out of the given options, acceleration is the only one that matches the criteria. Velocity is the rate of change of displacement, and displacement is the change in position. Therefore, acceleration is the only option that represents the rate of change of velocity.Rate this question:
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- 7.
Below is the expression for:
- A.
Acceleration
- B.
Velocity
- C.
Displacement
- D.
None of the above
Correct Answer
C. DisplacementExplanation
Displacement refers to the change in position of an object from its initial point to its final point. It is a vector quantity that takes into account both the magnitude and direction of the change in position. Acceleration and velocity are related to the rate of change of displacement, but they are not the same as displacement itself. Therefore, the correct answer is displacement.Rate this question:
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- 8.
Below is the expression for:
- A.
Acceleration
- B.
Velocity
- C.
Displacement
- D.
None of the above
Correct Answer
B. VelocityExplanation
Velocity is the correct answer because acceleration is the rate of change of velocity, and displacement is the change in position of an object. Therefore, neither acceleration nor displacement can be the expression for velocity. Velocity, on the other hand, is the rate at which an object changes its position in a particular direction, making it the appropriate expression for velocity.Rate this question:
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- 9.
A 2 kg mass is placed on the end of a spring with a constant 250 N/m. What will be the period of oscillation when the spring performs a simple harmonic motion?
- A.
0.56 s
- B.
0.28 s
- C.
0.089 s
- D.
70.2 s
Correct Answer
A. 0.56 sExplanation
The period of oscillation for a mass-spring system is determined by the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. In this case, the mass is 2 kg and the spring constant is 250 N/m. Plugging these values into the equation, we get T = 2π√(2/250) ≈ 0.56 s.Rate this question:
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- 10.
A 4 kg mass attached to a spring undergoes simple harmonic motion with a period 2.5 s. What is the spring constant?
- A.
25 N/m
- B.
35 N/m
- C.
250 N/m
- D.
150 N/m
Correct Answer
A. 25 N/mExplanation
The period of a mass-spring system is given by the equation T = 2π√(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we can solve for k: k = (4π²m)/T². Plugging in the given values, we get k = (4π² * 4) / (2.5)² ≈ 25 N/m. Therefore, the spring constant is 25 N/m.Rate this question:
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- 11.
A 5 kg mass undergoes simple harmonic motion on spring with a constant 70 N/m. If the amplitude of the oscillations is 3 m, what will the maximum speed be?
- A.
11.2 m/s
- B.
10.3 m/s
- C.
14.6 m/s
- D.
17.1 m/s
Correct Answer
A. 11.2 m/sExplanation
The maximum speed of an object undergoing simple harmonic motion can be calculated using the formula v_max = Aω, where A is the amplitude of the oscillations and ω is the angular frequency. The angular frequency can be calculated using the formula ω = √(k/m), where k is the spring constant and m is the mass of the object. In this case, the mass is 5 kg and the spring constant is 70 N/m. Plugging in these values, we get ω = √(70/5) = √14. The maximum speed is then v_max = 3 * √14 = 11.2 m/s.Rate this question:
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- 12.
If the following is a graph of the diplacement of an object at time t, which graph shows the velocity?
- A.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- B.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- C.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
Correct Answer
C.Explanation
The graph that shows the velocity would be the derivative of the displacement graph. Velocity is the rate at which the displacement changes with respect to time. Taking the derivative of the displacement graph will give us the velocity graph, which shows how the displacement is changing over time.Rate this question:
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- 13.
If the following is a graph of the diplacement of an object at time t, which graph shows the acceleration?
- A.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- B.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- C.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
Correct Answer
C.Explanation
The graph that shows the acceleration would be the graph of the slope of the displacement graph with respect to time. This is because acceleration is defined as the rate of change of velocity with respect to time, and velocity is the rate of change of displacement with respect to time. Therefore, the slope of the displacement graph represents the velocity, and the slope of the velocity graph represents the acceleration.Rate this question:
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- 14.
If the following is a graph of the force on an object at time t, which graph shows the displacement?
- A.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- B.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
- C.
![11th Grade Physics: Quiz On Simple Harmonic Motion! - Quiz]()
Correct Answer
C.Explanation
The graph of displacement would be a straight line with a positive slope. This is because displacement is the change in position of an object, and a positive slope indicates that the object is moving in a positive direction.Rate this question:
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- 15.
An oscillator completes 30 cycles in 15 seconds, what is its period?
- A.
0.25 seconds
- B.
0.50 seconds
- C.
1 second
- D.
2 seconds
Correct Answer
A. 0.25 secondsExplanation
The period of an oscillator is the time it takes to complete one cycle. In this case, the oscillator completes 30 cycles in 15 seconds. To find the period, we divide the total time by the number of cycles, which gives us 15 seconds divided by 30 cycles. Simplifying this, we get 0.5 seconds per cycle. Therefore, the period of the oscillator is 0.5 seconds, which is equivalent to 0.25 seconds.Rate this question:
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