Lesson 3.6.1.3
3.6.1.3 Simple harmonic systems Quiz: AQA Physics, Unit 6
20 questions
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Lesson 3.6.1.3, Simple harmonic systems: 20 multiple choice questions for the AQA Physics (7408), Unit 6: Further mechanics and thermal physics (A-level only), written with Revision Ninja.
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The 20 questions
-
The period of a mass-spring system is:
- T = 2 pi sqrt(m k)
- T = 2 pi m k
- T = 2 pi sqrt(m / k)
- T = 2 pi sqrt(k / m)
-
The period of a simple pendulum for small oscillations is:
- T = 2 pi l / g
- T = 2 pi sqrt(g / l)
- T = 2 pi l g
- T = 2 pi sqrt(l / g)
-
What is the small-angle approximation for sin(θ) when θ is in radians?
- Equal to cos(θ)
- Equal to θ
- Equal to 1
- Equal to tan(θ)
-
Which variable does not affect the time period of a simple pendulum?
- Gravitational field strength
- Length of string
- Pendulum length
- Mass of bob
-
What happens to the total mechanical energy of an undamped simple harmonic oscillator?
- Remains constant
- Increases continuously
- Decreases exponentially
- Fluctuates sinusoidally
-
What happens to the amplitude of a lightly damped oscillator over time?
- Remains constant
- Drops to zero
- Increases rapidly
- Decreases gradually
-
What describes the movement of an oscillator undergoing heavy damping?
- Resonant oscillations occur
- Constant amplitude oscillations
- Increasing amplitude oscillations
- No oscillations occur
-
A 0.20 kg mass on a spring of stiffness 50 N m^-1 oscillates. What is its period?
- 0.20 s
- 0.063 s
- 2.5 s
- 0.40 s
-
A simple pendulum has length 1.0 m (g = 9.8 m/s^2). What is its period?
- 4.0 s
- 0.32 s
- 1.0 s
- 2.0 s
-
By what factor must a simple pendulum's length change to double its period?
- Quarter
- Half
- Two
- Four
-
A mass-spring system has period 1.5 s with a 0.40 kg mass. What is the spring constant?
- 14 N m^-1
- 0.9 N m^-1
- 7.0 N m^-1
- 2.4 N m^-1
-
A simple pendulum of length 0.25 m is used where g = 1.6 m s^-2. What is its period?
- 6.2 s
- 1.0 s
- 0.40 s
- 2.5 s
-
How does the period of a mass-spring system change if mass is quadrupled?
- Halves
- Quadruples
- Doubles
- Remains unchanged
-
A 0.20 kg mass on a spring of stiffness 80 N m^-1 oscillates with amplitude 0.050 m. What is its total energy?
- 0.10 J
- 0.20 J
- 0.40 J
- 0.02 J
-
A 0.20 kg mass on a spring of stiffness 400 N m^-1 oscillates with amplitude 0.050 m. What is its maximum speed?
- 0.20 m s^-1
- 1.0 m s^-1
- 0.40 m s^-1
- 2.2 m s^-1
-
A student measures the period of a 0.50 m pendulum as 1.42 s. Using T = 2 pi sqrt(l/g), what is g?
- about 9.8 m s^-2
- about 0.98 m s^-2
- about 3.5 m s^-2
- about 19.6 m s^-2
-
A 0.50 kg mass on a spring has period 0.80 s. A further 0.20 kg is added to the mass. What is the new period?
- 0.67 s
- 0.95 s
- 0.80 s
- 1.13 s
-
What approximation is required for a simple pendulum to execute simple harmonic motion?
- Large angle approximation
- Zero friction approximation
- Constant speed approximation
- Small angle approximation
-
How does light damping affect the time period of a simple harmonic oscillator?
- Virtually unchanged
- Reduces to zero
- Doubles period
- Halves period
-
A 0.50 kg mass on a spring of stiffness 20 N m^-1 has amplitude 0.10 m. What is its maximum speed?
- 0.63 m s^-1
- 0.04 m s^-1
- 1.26 m s^-1
- 0.20 m s^-1
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