Spring-Block Oscillator

The Spring-mass Oscillator Experiment

A simple model of a system that exhibits Simple Harmonic Motion (SHM) is the spring-block oscillator, also called a linear simple harmonic oscillator. As shown below, this consists of a horizontal spring connected to a mass \( m \) that slides on an imaginary frictionless surface. The oscillator is described as linear because the force produced by the spring is directly proportional to \( x \).

Diagram of linear simple harmonic oscillator experiment using a spring connected to a mass. In its equilibrium position, the displacement of the block is 0, while the maximum displacement A at the oscillation extremes is also marked. StudySmarter Originals

From studying the kinematics of simple harmonic motion, we have equations for the displacement, velocity and acceleration of the oscillating block:

$$\text{Displacement } \quad x(t)=A\cos\left( (\omega t)+ \phi \right)$$

$$\text{Veloctiy} \quad v(t)=-A\omega\sin\left( (\omega t)+ \phi \right)$$

$$\text{Acceleration } \quad a(t)=-A\omega^2\cos\left( (\omega t)+ \phi \right)$$

In these kinematic equations for SHM, the variables are as follows:

• \( A \) is the maximum amplitude of the oscillation in meters \( \mathrm{(m)} \).

• \( \omega \) is the angular velocity of the SHM oscillation, given in radians-per-second \( \mathrm{(rad/s)} \).

• \( t \) is the time (instant the values are to be calculated at), measured in seconds \( \mathrm{(s)} \).

• \( \phi \) is the phase constant of the oscillation, which is a value representing the position of the oscillation at \( t=0\;\mathrm{s} \). It is given in radians.

Spring-mass Oscillator Formula

From Newton's second law, we know that force on an object is equal to its mass multiplied by its acceleration due to the force:

$$F=ma$$

And therefore the spring force can be equated to the mass acceleration:

$$F=(-k)x=ma$$

$$F=ma=(-m)A\omega^2\cos\left( (\omega t )+\phi \right)$$

$$F = (-k)x=(-k) A\cos\left( (\omega t) + \phi \right)$$

These can then be equated to obtain:

$$(-m)A\omega^2\cos\left( (\omega t )+\phi \right)=(-k) A\cos\left( (\omega t) + \phi \right)$$

The \( A \) and \( \cos\left( (\omega t )+\phi \right) \) factors in this equation cancel out together with a \( -1 \) factor, leaving us with:

$$m\omega^2=k$$

$$\omega = \sqrt{\dfrac{k}{m}}$$

This equation tells us the natural angular frequency for a spring-block oscillator system, which we can see is independent of the oscillation amplitude. The units of angular frequency are radians-per-second, with a rotation of \( 2\pi \) radians representing one full oscillation. We can also use this to find out the frequency and time period of the oscillation. This is because the angular frequency is directly related to the oscillation frequency and time period:

$$f= \frac{\omega}{2\pi}$$

$$T=\frac{1}{f}=2\pi\omega$$

Where the oscillation frequency \( f \) has units of Hertz \( \mathrm{(Hz)} \), and time period \( T \) is measured in seconds \( \mathrm{(s)} \).

This means that for the spring-block oscillator:

$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$

$$T=2\pi\sqrt{\frac{m}{k}}$$

Energy Equation for Oscillation of a Spring-Block

Examining how energy is transferred in the spring-block oscillator is useful to fully understand what is going on in the system, and why it exhibits SHM. The potential energy in a compressed or extended spring can be calculated considering the work needed to compress or extend the spring to that position:

$$W = \bar{F}x$$

Remember that spring force varies as it is compressed —but since it varies linearly, we can use an average force and get the same result. The average force \( \bar{F} \)is given by:

$$\bar{F}=\frac{1}{2}\left( F_1+F_0\right)$$

Where \( F_0 \) and \( F_1 \) are the spring force (in Newtons) produced by the uncompressed and compressed spring, respectively.

The distance \( x \) that the spring is compressed (or stretched) is the difference between the compressed \( x_1 \) and uncompressed \( x_0 \) displacements:

$$x = x_1-x_0$$

We can substitute in Hooke’s law for the spring forces\( F_1 \) and \( F_0 \) , giving us:

$$W=\bar{F}s=\left[ \frac{1}{2}(F_1+F_0) \right]x =\left[ \frac{1}{2}(kx_1+kx_0) \right]x $$

We can simplify this equation by expanding the terms:

$$W=-\frac{1}{2}\left( kx_1x+kx_0x \right)$$

As the uncompressed displacement \( x_0=0\;\mathrm{m} \), this implies that \( x_1=x \). This also means that all terms containing \( x_0 \) reduce to zero. Applying these to the equation for work done to compress the spring, we get:

$$PE_{\text{spring}}=\frac{1}{2}kx^2$$

We know from studying the kinematics of SHM that the oscillation velocity is highest when it passes through the equilibrium point and zero at the displacement extreme points. As we ignore energy losses when studying SHM, we know that all the potential energy stored in the spring when it is stretched or compressed must be transferred into the kinetic energy of the oscillation when the spring is uncompressed, and that the total energy \( PE+KE \) in the system is constant.

$$PE_0+KE_0=PE_1+KE_1$$

$$\frac{1}{2}kx_0^2+\frac{1}{2}mv_0^2+\frac{1}{2}kx_1^2+\frac{1}{2}mv_1^2$$

We can use this relationship to predict the maximum oscillation velocity when a spring is stretched by a certain amount. As the velocity \( v_1=0 \) at the extreme displacement point \( x_1 \) , and the displacement \( x_0=0 \) as the oscillation passes through the equilibrium point, some terms can be removed, leaving:

$$\frac{1}{2}kx_1^2=\frac{1}{2}mv_0^2$$

$$kx_1^2=mv_0^2$$

$$v_0=\sqrt{\frac{kx_1^2}{m}}$$

Let's put this into practice.

Vertical Spring-mass Oscillator

If we adjust the spring-block experiment so that the block now hangs vertically from the spring, as shown in the diagram below, this introduces the effect of gravity. Luckily, we can use a similar energy analysis approach to understand the behavior of this system.

The total energy in the system is still constant for a vertical spring-mass oscillator. Then, we just need to add a term for gravitational potential energy to our energy equation:

$$PE_0+KE_0=PE_1+KE_1$$

$$\frac12kx_0^2+mgx_0+\frac12mv_0^2=\frac12kx_1^2+mgx_1+\frac12mv_1^2$$