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AP Physics 1

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Unit 6

6.1 Period of Simple Harmonic Oscillators

6 min read•january 2, 2023

Daniella Garcia-Loos

Kanya Shah

Daniella Garcia-Loos

Kanya Shah

AP Physics 1 🎡

257 resources

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Simple Harmonic Motion exists whenever an object is being pulled towards an equilibrium point by a force that is proportional to the displacement from the equilibrium point. Two common examples of SHM are masses on a spring (one that obeys Hooke’s Law) and a pendulum (with a small angle displacement)

Newton's Second Law

Classically, the acceleration of an object interacting with other objects can be predicted using F = ma.

To use Newton's second law to solve a harmonic oscillator problem, you can follow these steps:

Identify the forces acting on the object in the system. These could include gravity, elastic forces, friction, and any other external forces.
Draw a free-body diagram to represent the forces acting on the object.
Determine the mass of the object and the acceleration it is experiencing.
Use Newton's second law to write an equation for the sum of the forces acting on the object. This equation is given by: F = ma, where F is the sum of the forces, m is the mass of the object, and a is the acceleration.
Substitute the known forces and the mass of the object into the equation and solve for the acceleration.
Use the acceleration to determine the velocity and position of the object as a function of time, using equations such as v = at and x = at^2/2.
Graph the velocity and position as a function of time to visualize the oscillatory motion of the object.
If necessary, use the equations for velocity and position to solve for any unknown quantities, such as the spring constant or the initial displacement of the object.
Repeat the process for any additional objects in the system, if applicable.

What is a Restoring Force?

Restoring forces can result in oscillatory motion. When a linear restoring force is exerted on an object displaced from an equilibrium position, the object will undergo a special type of motion called simple harmonic motion

Here are some key points about restoring forces:

A restoring force is a force that acts to bring an object back to its equilibrium position or to maintain its equilibrium position.
Restoring forces are often encountered in systems that experience oscillatory motion, such as a pendulum swinging back and forth or a mass on a spring oscillating up and down.
Restoring forces are often described as being "opposite in direction" to the displacement of the object from its equilibrium position. For example, if a mass on a spring is displaced to the right of its equilibrium position, the restoring force will be to the left, and vice versa.
Restoring forces can be caused by various factors, such as gravity, elastic forces, and friction.
The strength of a restoring force is often described by a spring constant, which is a measure of the stiffness of the spring or other force-generating element in the system. The higher the spring constant, the stronger the restoring force will be.
The restoring force can be calculated using the formula: F = -kx, where F is the restoring force, k is the spring constant, and x is the displacement of the object from its equilibrium position.
The presence of a restoring force is often used to explain why certain systems exhibit periodic motion, such as a mass on a spring oscillating up and down or a pendulum swinging back and forth.

** Note - for AP 1, we can assume that all the springs used are ideal springs. If you plan on taking the AP C: Mech exam, that will not be the case

Periods of a Pendulum and Spring

Amplitude is the height of the motion measured from the equilibrium point.

Period is the time that it takes for an object to complete one full cycle of its motion. The period is measured in seconds and is the inverse of the frequency (measured in Hz).

In the above equation, 𝜔 is the angular velocity of the object. This will be covered in detail in Unit 7: Torque & Rotational Motion

For a pendulum, the period of the oscillation can be described using the equation:

Where L is the length of the pendulum, and g is the acceleration due to gravity.

Looking at the equation, we can see that the period is proportional to the square root of the length. So a shorter pendulum will have a shorter period, and vice versa. In fact in order to double the period, we’d need to quadruple the length.

The period for a mass on a spring has a very similar equation. The only main difference is that the spring’s period doesn’t depend on length and acceleration due to gravity, but rather the mass hung on the spring and the spring constant. For a more in-depth derivation, check out this link.

Key Features of Pendulums & Springs

When dealing with pendulums and springs, a lot of the questions you’ll be dealing with refer to the velocities, forces, and accelerations at various locations in the oscillation:

The net force and acceleration vectors always point in the same direction.
The force and acceleration vectors are greatest when the spring is fully stretched and compressed
The velocity vector is 0 at the extremes where the spring is fully stretched and compressed.

The velocity vector is at its maximum when the mass passes through the equilibrium point. (This is also where the force and acceleration vectors are 0)

Example Problem 1:

A mass of 1 kg is attached to a spring with a spring constant of 50 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.2 meters from its equilibrium position and released from rest. What is the period of the oscillation?

Solution:

The period of an oscillation is the time it takes for the oscillating object to complete one full oscillation. For a simple harmonic oscillator, the period is given by the formula: T = 2pisqrt(m/k), where T is the period, m is the mass of the object, k is the spring constant, and pi is a constant equal to 3.14.

In this problem, the mass of the object is 1 kg, the spring constant is 50 N/m, and pi is 3.14.

Therefore, the period of the oscillation is: T = 23.14sqrt(1 kg / 50 N/m) = 0.63 seconds

This means that the period of the oscillation is 0.63 seconds.

Example Problem 2:

A mass of 2 kg is attached to a spring with a spring constant of 100 N/m and is allowed to oscillate vertically in a frictionless environment. The mass is initially displaced 0.5 meters from its equilibrium position and released from rest. What is the period of the oscillation?

Solution:

In this problem, the mass of the object is 2 kg, the spring constant is 100 N/m, and pi is 3.14.

Therefore, the period of the oscillation is: T = 23.14sqrt(2 kg / 100 N/m) = 0.44 seconds

This means that the period of the oscillation is 0.44 seconds.

🎥Watch: AP Physics 1 - Problem Solving q +a Simple Harmonic Oscillators

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