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VCE (Victoria) Oscillatory Motion and Waves — practice questions

Explore 54 free Oscillatory Motion and Waves questions in VCE (Victoria) Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Oscillatory Motion and Waves questions and exercises with answers

Showing questions 1–30 of 54.

In Hooke's law, the force constant k measures how hard a system is to deform. What are its SI units?Read the answer and worked solution →A car body settles 1.50 cm when a 90.0 kg passenger climbs in. Taking g as 9.80 m/s squared, what is the force constant of the suspension?Read the answer and worked solution →A spring with a force constant of 250 N/m is compressed 4.00 cm from its natural length. How much elastic potential energy is stored in it?Read the answer and worked solution →A spring scale stretches 5.00 cm when a 2.00 kg fish hangs at rest from it. With g taken as 9.80 m/s squared, what is the force constant of its spring?Read the answer and worked solution →A system completes 5 full oscillations every second. How is that frequency written in SI units?Read the answer and worked solution →Middle C on a typical musical instrument has a frequency of 264 Hz. How long does one complete oscillation take?Read the answer and worked solution →A medical imaging device produces ultrasound by oscillating with a period of 0.400 microseconds. What is the frequency of that oscillation?Read the answer and worked solution →Frequency and period are linked by f = 1 / T. If the period of an oscillation is halved, what happens to its frequency?Read the answer and worked solution →During strenuous exercise a heart beats 150 times per minute. What is the time taken for one beat?Read the answer and worked solution →In describing oscillations, what is meant by one cycle?Read the answer and worked solution →Mains electrical power in some countries alternates at 60.0 Hz. What is the period of that oscillation?Read the answer and worked solution →What does the amplitude of an oscillation measure?Read the answer and worked solution →Which pair of quantities are the only ones that set the period and frequency of a simple harmonic oscillator?Read the answer and worked solution →A 0.400 kg block on a spring of force constant 100 N/m oscillates freely on a frictionless surface. What is the period of the motion?Read the answer and worked solution →A 900 kg car rests on a suspension of force constant 6.53 x 10^4 N/m and bounces freely on it. What are the frequency and period of the bounce?Read the answer and worked solution →A pen fixed to an object bouncing on a spring draws on a paper strip pulled steadily sideways. What shape does the trace take?Read the answer and worked solution →What counts as a simple pendulum?Read the answer and worked solution →Below roughly what swing angle does a simple pendulum behave as a simple harmonic oscillator?Read the answer and worked solution →What is the period of a simple pendulum 1.00 m long at a place where g is 9.80 m/s squared?Read the answer and worked solution →A pendulum is to be built with a period of 2.00 s at a place where g is 9.80 m/s squared. What length of string is needed?Read the answer and worked solution →A simple pendulum of length 1.0000 m is timed and found to have a period of 2.0100 s. What is the acceleration due to gravity at that place?Read the answer and worked solution →For an undamped simple harmonic oscillator, which quantity stays constant throughout the motion?Read the answer and worked solution →Which change would increase the maximum speed reached by a mass oscillating on a spring?Read the answer and worked solution →A 0.500 kg block on a spring of force constant 200 N/m oscillates with an amplitude of 0.100 m. What is its maximum speed?Read the answer and worked solution →A point goes once round a circle in a time T. What is its angular velocity, in radians per unit time?Read the answer and worked solution →What is critical damping in an oscillating system?Read the answer and worked solution →How does an overdamped system compare with a critically damped one on its way back to equilibrium?Read the answer and worked solution →A 0.200 kg object slides on a horizontal surface where the coefficient of kinetic friction is 0.0800. Taking g as 9.80 m/s squared, what is the frictional force on it?Read the answer and worked solution →What is meant by the natural frequency of a system?Read the answer and worked solution →When you tune a radio to a particular station, what are you actually adjusting?Read the answer and worked solution →

Worked practice questions

In Hooke's law, the force constant k measures how hard a system is to deform. What are its SI units?

  1. Joules per meter, written J/m
  2. Newtons per meter, written N/m
  3. Newtons per second, written N/s
  4. Meters per newton, written m/N
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Correct answer: B. Newtons per meter, written N/m

1. Rearranging F = -k x gives k = F / x, so k carries the units of force divided by the units of length. 2. Force is measured in newtons and displacement in meters, so k is measured in newtons per meter. 3. The units of k are newtons per meter (N/m). 4. Joules per meter would be an energy gradient and meters per newton is the reciprocal of a force constant, so neither fits. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.1 Hooke's Law: Stress and Strain Revisited_

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A car body settles 1.50 cm when a 90.0 kg passenger climbs in. Taking g as 9.80 m/s squared, what is the force constant of the suspension?

  1. 1.35 x 10^3 N/m
  2. 5.88 x 10^2 N/m
  3. 5.88 x 10^4 N/m
  4. 8.82 x 10^4 N/m
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Correct answer: C. 5.88 x 10^4 N/m

1. While the car sits still, the restoring force from the springs equals the passenger's weight. 2. Weight w = m g = (90.0 kg)(9.80 m/s squared) = 882 N. 3. Convert the settling distance: 1.50 cm = 1.50 x 10^-2 m. 4. From k = F / x, k = 882 N / (1.50 x 10^-2 m) = 5.88 x 10^4 N/m. 5. Dividing 882 N by 1.50 without converting to meters gives 5.88 x 10^2 N/m, which is the trap in one option. 6. Multiplying weight by displacement instead of dividing gives 13.2 N m, an energy, not a force constant. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section 16.1 Hooke's Law: Stress and Strain Revisited_

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A spring with a force constant of 250 N/m is compressed 4.00 cm from its natural length. How much elastic potential energy is stored in it?

  1. 0.0500 J
  2. 5.00 J
  3. 10.0 J
  4. 0.200 J
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Correct answer: D. 0.200 J

1. Elastic potential energy stored in a Hooke's law system is PE = (1/2) k x squared. 2. Convert the compression: 4.00 cm = 0.0400 m. 3. Square it: (0.0400 m) squared = 1.60 x 10^-3 m squared. 4. PE = (1/2)(250 N/m)(1.60 x 10^-3 m squared) = 0.200 J. 5. Forgetting to square the displacement gives 5.00 J, and dropping the factor of one half gives 0.400 J, so both shortcuts mislead. _Source: OpenStax College Physics (CC BY 4.0), Ch 16 "Oscillatory Motion and Waves", section Energy in Hooke's Law of Deformation_

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