If you've ever seen footage of astronauts aboard the international space station (ISS), you've probably seen how much effort they put into exercising and keeping fit while in space. But how can they keep track of their weight, given that objects in orbit experience weightlessness? The solution is a specialized weighing device that uses the principles of Simple Harmonic Motion (SHM) to determine an astronaut's mass. By attaching themselves to a spring with a known stiffness, the natural frequency of the astronaut-spring system oscillations can be measured and used to calculate their mass.
The Spring-Block Oscillator is a classic physics experiment that demonstrates the same principles that the scales on the ISS utilizes. This article outlines the spring-block experiment, derives an equation to find the natural frequency and time period of different oscillators, and introduces an energy analysis method for finding the velocity and displacement at different points in an oscillation.
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:
$$F=(-k)x=ma$$
This gives us equations for the spring force in terms of acceleration \( a \) and displacement \( x \). We can substitute the equations for SHM displacement and acceleration into these force equations, which gives:
$$F=ma=(-m)A\omega^2\cos\left( (\omega t )+\phi \right)$$
$$F = (-k)x=(-k) A\cos\left( (\omega t) + \phi \right)$$
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:
$$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$$
$$T=2\pi\sqrt{\frac{m}{k}}$$
In the simple linear harmonic oscillator shown below, the spring has a spring constant of \( 1000\;\mathrm{N/m} \) and the block has a mass of \( 5\;\mathrm{kg} \). Determine the frequency and time period of the system’s oscillations.
Illustration of a simple linear harmonic oscillator consisting of a spring and a block. StudySmarter Originals
We can find the angular frequency using the equation we derived earlier:
$$\omega = \sqrt{\dfrac{k}{m}} = \sqrt{\dfrac{1000\;\mathrm{N/m}}{5\;\mathrm{kg}}}= 14.14\; \mathrm{rad/s}$$
Since one full oscillation is \( 2\pi \) radians, we can divide the angular frequency by \( 2\pi \) to find the frequency in Hertz:
$$f = \frac{\omega}{2\pi}=\frac{14.14\;\mathrm{rad/s}}{2\pi}=2.25\;\mathrm{Hz}$$
Alternatively, we can calculate this directly using our equation for frequency:
$$f= \frac{1}{2\pi}\sqrt{\frac{k}{m}}=\frac{1}{2\pi}\sqrt{\frac{ 1000\;\mathrm{N/m}}{5\;\mathrm{kg}}}=2.25\;\mathrm{Hz}$$
The time period can also be calculated directly:
$$T=2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{5\;\mathrm{kg}}{1000\;\mathrm{N/m}}}=0.44s$$
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.
A spring-block oscillator experiment is shown below. If the spring is initially compressed by \( 10\;\mathrm{cm} =0.01\;\mathrm{m} \), how fast will the block be moving as it passes through the equilibrium point after it is released?
Illustration of a simple linear harmonic oscillator consisting of a spring and a block. In this diagram, the block is displaced to the left. StudySmarter Originals
Let's substitute the known data into the equation for velocity and simplify.
$$v_0 = \sqrt{\dfrac{kx_1^2}{m}}=\sqrt{\dfrac{400\;\mathrm{N/m}\times 0.1\;\mathrm{m}^2}{2\;\mathrm{kg}}}=\sqrt{\dfrac{4}{2}}=\sqrt{2}=1.414\;\mathrm{m/s}$$
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:
$$PE_0+KE_0=PE_1+KE_1$$
$$\frac12kx_0^2+mgx_0+\frac12mv_0^2=\frac12kx_1^2+mgx_1+\frac12mv_1^2$$
In this example, we treat acceleration due to gravity as \( g = 10\;\mathrm{N/kg} \).
The experiment setup below shows a \( m=3\;\mathrm{kg} \) block hanging from a spring with a stiffness (spring constant) of \( k=600\;\mathrm{N/m} \). Before the block is attached, the spring rests unstretched with the lowest point extending to \( x_0=0\; \mathrm{m} \). When the block is attached, the spring stretches until the spring force cancels the gravitational force and the system is in equilibrium at displacement \( x_1 \). Find the value of \( x_1 \).
A vertical spring-block oscillator experiment, showing the unstretched spring at displacement \( x_0 \), the equilibrium position of the loaded spring at \( x_1 \) and the maximum extension of the spring at \( x_2 \). StudySmarter Originals
The equilibrium position of the system can be found by calculating the displacement required for gravity and the spring force to equate:
$$W = mg = 3\;\mathrm{kg}\times10\;\mathrm{N/kg}=10\;\mathrm{N}$$
$$\therefore F_{\text{spring}}=-kx_1=-600\;\mathrm{N/m}\times x_1\;\mathrm{m}=30\;\mathrm{N}$$
$$x_1=\frac{ 30\;\mathrm{N}}{-600\;\mathrm{N/m}}=-0.05\;\mathrm{m}$$
If the block is pulled downwards so that the spring extends another \( 0.15\; \mathrm{m} \) to\( x_2 =-0.2\;\mathrm{m} \). What will be the maximum velocity and displacement of the block oscillation after it is released?
At the stretched position we can calculate the total energy in the system:
$$\text{Total Energy}= \frac{ 1}{2}kx_2^2+mgx_2+\frac{ 1}{2}mv_2^2$$
As this is an extreme point, the velocity is zero and the kinetic term can be removed: $$\text{Total Energy}= \frac{ 1}{2}kx_2^2+mgx_2$$
Now let's substitute the know values and simplify.
$$\text{Total Energy at }x_2=\frac{1}{2}\times600\;\mathrm{N/m}(-0.2\;\mathrm{m})^2+3\;\mathrm{kg\times10\;\mathrm{N/kg}\times-0.2\;\mathrm{m}}$$
$$\text{Total Energy at }x_2=12 J$$
When the block is released, the spring force accelerates it upwards. We know that the oscillation velocity is greatest when it passes through the equilibrium point. We already calculated the total system energy of the system to be \( 12\;\mathrm{J} \), so we can use it to find the velocity \( v_1 \) at \( x_1=-0.05\;\mathrm{m} \):
$$12\;\mathrm{J} = \frac{1}{2}kx_1^2+mgx_1+\frac{1}{2}mv_1^ 2$$
$$12\;\mathrm{J} =\left(\frac12\times600\;\mathrm N/\mathrm m\times{(-0.05\;\mathrm m)}^2\right)+\left(3\;\mathrm{Kg}\times10\;\mathrm N/\mathrm{kg}\times-0.05\;\mathrm m\right)+\left(\frac12\times3\;\mathrm{kg}\times v_1^2\right)$$
$$12\;\mathrm{J} =0.75\;\mathrm{J}-1.5\mathrm{J}+1.5\;\mathrm{kg}\times v_1^2$$
$$ \sqrt{\frac{12.75\;\mathrm{J}}{1.5\;\mathrm{kg}}}=v_1 = 2.916\;\mathrm{m/s} $$
We could also use this energy method to find the maximum displacement of the oscillation, but there is a more straightforward option —we know that a key feature of the simple harmonic motion is that the displacement magnitude in either direction from the equilibrium point is equal!
In this example, our minimum displacement is at \( x_2=-0.2\;\mathrm{m} \), and the equilibrium point is at \( x_1=-0.05\;\mathrm{m} \). This means the oscillation amplitude \( A=0.15\;\mathrm{m} \). We can add the amplitude to the equilibrium point to find the maximum displacement of the oscillation:
$$x_1+A = -0.05\;\mathrm{m}+0.15\;\mathrm{m}=0.1\;\mathrm{m}$$
The Spring-Block Oscillator - Key takeaways
- A Spring-block oscillator is a system comprising of a spring attached to a block/mass that slides on a frictionless surface. The block will oscillate around an equilibrium position with simple harmonic motion. This experiment is also known as a linear simple harmonic oscillator.
- The spring produces a restoring force on the block that is inversely proportional to its displacement, a key feature of SHM.
- The relationship between spring force and extension is defined by Hooke's law \( F=-kx \) while the spring remains within its elastic limit.
- We can find the natural angular frequency, frequency and time period of a spring-block oscillator using the equations: \( \omega = \sqrt{\dfrac{k}{m}} \),\( f = \dfrac{1}{2\pi} \sqrt{\dfrac{k}{m}}\) & \( T=2\pi\sqrt{\dfrac{m}{k}} \), respectively.
- As we ignore energy losses when analyzing SHM, the total energy in the spring-block oscillator system (sum of the kinetic energy, spring potential energy, and gravitational potential energy) must be constant at any point: \( PE_0+KE_0 = PE_1+KE_1 \).
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