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Periodic Motion

Periodic motion is a motion that repeats at certain time intervals, such as a rocking chair moving back and forth. Another example would be a swinging pendulum. The time it takes for the pendulum to complete one oscillation is called the period T’, which is usually measured in secondsThe term ‘period’ is used to describe the time it takes for an event to happen, regardless of whether that event is repetitive or not.

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Periodic motion is a motion that repeats at certain time intervals, such as a rocking chair moving back and forth. Another example would be a swinging pendulum. The time it takes for the pendulum to complete one oscillation is called the period T’, which is usually measured in seconds. The term ‘period’ is used to describe the time it takes for an event to happen, regardless of whether that event is repetitive or not.

However, we are here concerned with repetitive events, as periodic motion is repetitive. We, therefore, also need to take into consideration the frequency ‘f . This specifies the number of times the event happens in one unit time (usually a second), and it is measured in Hertz.

All events that repeat after a certain amount of time (period) are said to be periodic, and all periodic motions have a frequency. The relationship between frequency and period is expressed in the following equation:

\[f = \frac{1}{T}\]

What are the types of periodic motion?

While periodic motion can be seen in a variety of events, it is helpful to distinguish between specific types of motion such as circular motion and simple harmonic motion.

Circular motion

Circular motion occurs around a central point with a constant radius and a certain speed. There are a variety of examples of circular motion in our daily lives, such as the blades of a ceiling fan rotating or a giant carousel wheel at the park.

While the circular motion is present, the object’s distance from the axis of rotation, also known as radius, stays constant at all times. The velocity of the object, however, changes continuously since velocity is a vector quantity that depends on both speed and direction.

The change in velocity indicates the presence of an acceleration, which is caused by a centripetal force. The centripetal force is the force that maintains the circular motion, acting towards the direction of the centre and along the radius.

\[F = \frac{m \cdot v^2}{r}\]

  • F is the centripetal force in Newtons.
  • m is the mass of the object in kg.
  • v is the velocity of the object in m/s.
  • r is the radius of the objects orbit.

Periodic Motion, Circular motion, StudySmarter

Fig. 1 - A diagram of circular motion

The rate of change of the object’s angular position is the angular velocity (ω). If an object is moving, as shown in Figure 1, in a circular motion whose radius is r, with the period of one rotation being T, we can determine its angular velocity using the formula below:

\[\frac{d \theta}{d t} = \omega \space rad/s = \frac{2 \pi}{T} = 2 \pi ƒ\]

Here, f is the frequency in Hertz.

The relation between the object’s speed (v) and its angular velocity can be expressed as follows:

\[v = \omega \cdot T\]

The angle covered in a time period (t ) is also dependent on the angular velocity (ω) and can be expressed as shown below (the values are given in radians).

\[\theta = \omega \cdot t\]

A tennis ball connected to a stick with a 1.2m long rope is following a circular motion around the stick at a constant velocity. The tennis ball has a mass of 60g and is moving at a velocity of 28.2 m/s. Calculate the centripetal force.

Solution:

You have learned that the equation for the centripetal force is:

\[F = \frac{m \cdot v^2}{r}\]

The length of the rope (r) is 1.2m, the mass of the ball (m) is 60g, and the velocity (v) is 28.2 m/s. Let’s put these variables into the equation to calculate F.

\[F = \frac{(0.06 \space kg) \cdot (28.2 \space m/s)^2}{1.2 \space m} = 39.7 \space N\]

Simple harmonic motion

Simple harmonic motion is the name given to a system that can be explained using Hooke’s law. This states that an object’s displacement or deformation size is directly proportional to the load or force applied to it. Any system obeying this law is known as a simple harmonic oscillator.

Hooke’s law states that the force it takes to extend or compress a spring is directly proportional to the distance the spring is going to extend or compress and the force constant k, which is a constant factor characteristic of the spring.

\[F[N] = k [N/m] \cdot x [m]\] (N: Newton, m: meter)

If there is no friction nor any other external force acting on the object, it will oscillate with equal displacement on either side of the position the oscillator takes when there are no forces acting on it.

Periodic Motion, Simple harmonic oscillator, StudySmarter

Fig. 2 - A diagram of a simple harmonic oscillator

To calculate the period or frequency of simple harmonic oscillators, we need to consider the object’s mass m and the force constant of the spring k, but we don’t need to consider the amplitude of the applied force, which is independent of the period.

For instance, regardless of whether you pluck the strings of a guitar hard or more softly, they will oscillate with the same frequency because the period is constant. The period, mass, and spring constant relationship are given by the formula below.

\[T = 2 \pi \cdot \sqrt{\frac{m}{k}}\]

  • m is the mass of the object in kg.
  • k is the force constant in N/m.
  • T is the oscillation period in seconds.

You can also express this using the frequency rather than the period. If the frequency f is equal to 1/T, we obtain the following expression:

\[ƒ = \frac{1}{2\pi} \cdot \sqrt{\frac{k}{m}}\]

A cube with a mass of 4kg is attached to a string with a force constant of 2 N/m. Calculate the period and the frequency of the cube if a force of 10N is applied to compress the string.

Solution:

The mass of the cube m, the force constant k, and the force F have all been specified. We need to remember that the period or the frequency of simple harmonic motion is independent of the force applied.

As we said, the equation for finding the period of simple harmonic motion is:

\[T = 2 \pi \cdot \sqrt{\frac{m}{k}}\]

Adding the known variables, we get:

\[T = 2 \pi \cdot \sqrt{\frac{4}{2}} = 2 \cdot 3.14 \cdot 1.41 = 8.85 \space s\]

To find the frequency of the motion, we can either use the equation for finding the frequency or take a shortcut by remembering that the frequency f is equal to 1/T.

\[ƒ = \frac{1}{T} = \frac{1}{8.85} = 0.113 \space Hz = 113 \space mHz\]

The simple pendulum

There are a variety of applications for pendulums in our daily lives. Some are very important, like the pendulums used in clocks, while others are for fun, such as a children’s swing in a playground. For small displacements, a simple pendulum can be considered a simple harmonic oscillator.

A simple pendulum is an object with a certain mass suspended from a string or wire.

Periodic Motion. Simple pendulum. StudySmarter

Fig. 3 - Diagram of a simple pendulum and the forces acting on it

As you can see in Figure 3, the linear displacement from the equilibrium is s, the gravitational force is w = mg, where m is mass and g is the acceleration due to gravity in m/s², while T is the tension force from the string that is keeping the object connected to the beam.

We can express the net force (F) acting on the pendulum as follows:

\[F = -mg \cdot \sin(\theta)\]

The relation between s and L is:

\[s = L \cdot \sin(\theta)\]

You can see from this equation that θ is equal to s/L. If we add this to the equation for finding the net force and take sin (θ) as sin (θ) ≈ θ, we can express it as follows:

\[F \approx -\frac{mg}{L} \cdot s\]

For small angles of oscillation (θ), the value of sin (θ) is approximately equal to θ itself.

This is another form of F = -kx, and it tells us that we can take k = mg/L when the displacement is x = s. This helps us to express the period in another way.

As we know, the period of a simple harmonic oscillator can be determined as follows:

\[T = 2 \pi \cdot \sqrt{\frac{m}{k}}\]

If we replace k with mg/L, we get:

\[T = 2 \pi \cdot \sqrt{\frac{m}{mg/L}}\]

Thus,

\[T = 2 \pi \cdot \sqrt{\frac{L}{g}}\]

Find the acceleration due to gravity for a simple pendulum with a length of 50 cm and a period of 1.4576 s.

Solution:

If we square the equation for finding the period and solve it for g, we can find the acceleration due to gravity.

As we know, the equation for finding the period is:

\[T = 2 \pi \cdot \sqrt{\frac{L}{g}}\]

If we square both sides and solve it for g, we get:

\(g = 4 \pi^2 \cdot \frac{L}{T^2}\)

\(g = 4 \pi^2 \cdot {0.5 \space m}{(1.4576 \space s)^2 = 9.281

Periodic Motion - Key takeaways

  • Periodic motion is a motion that repeats itself at certain time intervals.
  • While periodic motion can be seen in a variety of events, we study periodic motion in more specific types of motion, such as circular motion and simple harmonic motion.
  • The period of simple harmonic motion is independent of the amplitude of the force being applied.

Frequently Asked Questions about Periodic Motion

No, periodic motions are not all simple harmonic motions. They can also be circular motions.

We can measure time by using periodic motion if we know the period of the motion, which is the time it takes for the motion to complete one cycle. After that, we can count the periods and multiply the counted value with the value of the period to measure time in seconds.

No, all periodic motions are not oscillatory, but all oscillatory motions are periodic motions because they repeat themselves at certain time intervals called periods.

Test your knowledge with multiple choice flashcards

What is the definition of a radian?

What units do we use for the time period in circular motion?

What units do we use for angular speed?

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