Instruments 5.2
By the end of this session, you will be able to:
If an object’s position is \(x(t)\):
Velocity: \[v(t) = \frac{dx(t)}{dt} \tag{16}\]
Acceleration: \[a(t) = \frac{dv(t)}{dt} = \frac{d^2 x(t)}{dt^2} \tag{17}\]
If we know acceleration \(a(t)\), we can integrate to get:
Velocity: \[v(t) = v(0) + \int_0^t a(\tau)\, d\tau \tag{18}\]
Position: \[x(t) = x(0) + \int_0^t v(\tau)\, d\tau \tag{19}\]
Tip
An accelerometer measures \(a(t)\), but with integration (usually electronic) we can estimate \(v(t)\) and \(x(t)\).
Often accelerations are normalized to earth gravity:
So:
An automobile is accelerating away from a stop sign at \(26.4\ \mathrm{ft/s^2}\).
Find the acceleration in:
Solution
Convert feet to meters:
\[ a = (26.4\ \mathrm{ft/s^2})(12\ \mathrm{in/ft})(2.54\ \mathrm{cm/in})(0.01\ \mathrm{m/cm}) \]
\[ a \approx 8.05\ \mathrm{m/s^2} \]
Express in g’s:
\[ a_{\mathbf{g}} = \frac{8.05\ \mathrm{m/s^2}}{9.8\ \mathrm{m/s^2}/\mathbf{g}} \approx 0.82\ \mathbf{g} \]
Note
That car launch is about \(82\%\) of gravity. Pretty sporty, but still under \(1\ \mathbf{g}\).
Four basic motion categories:
Sensor Matching

For simple periodic (harmonic) vibration:
\[ x(t) = x_0 \sin(\omega t) \tag{20} \]
Where:

Relationship between frequency and angular frequency:
\[ \omega = 2\pi f \tag{21} \]
Starting with \[x(t) = x_0 \sin(\omega t)\]
Velocity (first derivative):
\[ v(t) = \omega x_0 \cos(\omega t) \tag{22} \]
Acceleration (second derivative):
\[ a(t) = -\omega^2 x_0 \sin(\omega t) \tag{23} \]
Peak acceleration:
\[ a_{\text{peak}} = \omega^2 x_0 \tag{24} \]
Important
Peak acceleration rises with \(\omega^2\). Doubling frequency quadruples peak acceleration (for the same displacement).
A water pipe vibrates at \(f = 10\ \mathrm{Hz}\) with displacement \(x_0 = 0.5\ \mathrm{cm}\).
Find:
Solution
Compute \(\omega\) and convert \(x_0\):
\[ a_{\text{peak}} = \omega^2 x_0 = (20\pi)^2 (0.005) \approx 19.7\ \mathrm{m/s^2} \]
\[ a_{\text{peak}} = \frac{19.7\ \mathrm{m/s^2}}{9.8\ \mathrm{m/s^2/g}} \approx 2.0\ \mathbf{g} \]
Warning
A 2‑g vibration is serious; repeated exposure can damage pipes, valves, and electronics.
Shock = very large deceleration over a short time.
Typical characteristics:

Key quantities:
A TV set is dropped from height \(h = 2\ \mathrm{m}\). If the shock duration at impact is \(T_d = 5\ \mathrm{ms}\), find the average shock in g.
Solution
\[ v^2 = 2(9.8)(2.0) \Rightarrow v \approx 6.3\ \mathrm{m/s} \]
\[ \bar{a} = \frac{v}{T_d} = \frac{6.3\ \mathrm{m/s}}{5 \times 10^{-3}\ \mathrm{s}} \approx 1260\ \mathrm{m/s^2} \]
\[ \bar{a} \approx \frac{1260\ \mathrm{m/s^2}}{9.8\ \mathrm{m/s^2/g}} \approx 128\ \mathbf{g} \]
Warning
\(128\ \mathbf{g}\) average shock explains why the TV breaks apart on impact.
Core idea: combine Newton’s law and Hooke’s law.
Equate forces:
\[ ma = k\Delta x \tag{25} \]
Solve for acceleration:
\[ a = \frac{k}{m} \Delta x \tag{26} \]
So:

Any mass–spring system has a natural frequency \(f_N\):
\[ f_N = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \tag{27} \]
And a damped transient response after a disturbance:
\[ X_T(t) = X_0 e^{-\alpha t} \sin(2\pi f_N t) \tag{28} \]
Where:

Note
Near \(f_N\), the system can resonate → large motion and nonlinear response.
Consider a spring–mass accelerometer mounted on a vibrating table.
Table motion: \[x_{\text{table}}(t) = x_0 \sin(\omega t)\]
Table acceleration: \[a(t) = -\omega^2 x_0 \sin(\omega t)\]
Using \(ma = k\Delta x\):
\[ \Delta x = -\frac{m x_0}{k}\, \omega^2 \sin(\omega t) \tag{29} \]
Where:
Key idea (ignoring resonance):


Rules of thumb:
An accelerometer:
Find:
Solution
\[ a = \frac{k}{m} \Delta x = \frac{3.0 \times 10^3}{0.05}(0.02) \approx 1200\ \mathrm{m/s^2} \]
Convert to g’s:
\[ a \approx \frac{1200}{9.8} \approx 122\ \mathbf{g} \]
\[ f_N = \frac{1}{2\pi} \sqrt{\frac{k}{m}} = \frac{1}{2\pi} \sqrt{\frac{3.0 \times 10^3}{0.05}} \approx 39\ \mathrm{Hz} \]
Tip
This accelerometer can measure up to about ±122 g, but has a fairly low natural frequency (~39 Hz), so it is not suitable for high‑frequency vibration.
We now examine several implementation types, all using the same basic mass–spring concept but with different displacement sensing:
Each has characteristic range, natural frequency, and output type.
Note
Advantages: simple, low cost. Disadvantages: limited bandwidth, mechanical wear, not suited for harsh vibration.
Use an LVDT (Linear Variable Differential Transformer) to measure mass displacement.

Typical features:
Key traits:
Important
These are dynamic‑only sensors: no output for constant acceleration (e.g., constant tilt).
Use the piezoelectric effect: a crystal generates a voltage when mechanically stressed.

Features:
For nonperiodic or slowly varying acceleration (e.g., vehicle motion, elevator):
Requirements:
Integration strategy:

An accelerometer outputs \(14\ \mathrm{mV/g}\).
Design signal conditioning to produce a velocity signal scaled at \(0.25\ \mathrm{V}\) per \((\mathrm{m/s})\). Determine:
Solution outline
\[ K = 14\ \frac{\mathrm{mV}}{\mathbf{g}} \cdot \frac{1\ \mathbf{g}}{9.8\ \mathrm{m/s^2}} \approx 1.43\ \frac{\mathrm{mV}}{\mathrm{m/s^2}} \]
\[ V_v = -\frac{1}{RC} \int V_a dt = -\frac{K}{RC} v \]
\[ V_{\text{out}} = \left(\frac{R_2}{R_1}\right)\frac{K}{RC} v \]
We want \(V_{\text{out}} = 0.25 v\), so:
\[ 0.25 = \left(\frac{R_2}{R_1}\right)\frac{K}{RC} \]
Choose \(R = 1\ \mathrm{M\Omega}\), \(C = 1\ \mu\mathrm{F}\) ⇒ \(RC = 1\).
Then:
\[ \frac{R_2}{R_1} = \frac{0.25}{1.43 \times 10^{-3}} \approx 175 \]
So one practical choice: \(R_1 = 1\ \mathrm{k\Omega}\), \(R_2 = 175\ \mathrm{k\Omega}\).
Key spec checklist:
Pressure = force per unit area exerted by a fluid (liquid or gas).
SI unit:
Often use prefixes:
Other common units:
Measurement systems must clearly state whether they are reading:
Atmospheric pressure at sea level: ~\(14.7\ \mathrm{psi}\) (1 atm).
Often, we care about pressure relative to atmosphere, not absolute.
Relationship:
\[ p_g = p_{\text{abs}} - p_{\text{at}} \tag{30} \]
Note
A sealed container at \(14.7\ \mathrm{psi}\) absolute, sitting at sea level, has \(p_g = 0\) (no net force on the walls).
For a liquid column of depth \(h\):
SI form:
\[ p = \rho g h \tag{31} \]
Where:
English form using weight density \(\rho_w\) \((\mathrm{lb/ft^3})\):
\[ p = \rho_w h \tag{32} \]
Where \(p\) is in \(\mathrm{lb/ft^2}\).
To get psi, divide by 144 (since \(1\ \mathrm{ft^2} = 144\ \mathrm{in^2}\)).
Tip
“Feet (or meters) of water” is a pressure unit: it’s shorthand for the pressure produced by that water column height.
A tank holds water with depth \(7.0\ \mathrm{ft}\).
Find the pressure at the bottom in:
Given: water density \(\rho = 10^3\ \mathrm{kg/m^3}\).
Solution
Convert depth:
\[ h = 7.0\ \mathrm{ft} = 7.0 \cdot 0.3048\ \mathrm{m/ft} \approx 2.1\ \mathrm{m} \]
Use Equation (31):
\[ p = \rho g h = (10^3)(9.8)(2.1) \approx 2.1 \times 10^4\ \mathrm{Pa} = 21\ \mathrm{kPa} \]
To express in psi, we can use \(\rho_w \approx 62.4\ \mathrm{lb/ft^3}\).
Pressure in \(\mathrm{lb/ft^2}\):
\[ p = \rho_w h = (62.4\ \mathrm{lb/ft^3})(7.0\ \mathrm{ft}) = 437\ \mathrm{lb/ft^2} \]
Convert to psi:
\[ p = \frac{437}{144}\ \mathrm{psi} \approx 3\ \mathrm{psi} \]
Most industrial pressure sensors first convert pressure → displacement, then measure displacement electrically.
Common mechanical elements:
Then convert displacement with:



A thin flexible metal diaphragm with pressures \(p_1\) and \(p_2\) on each side.
Net force:
\[ F = (p_2 - p_1) A \tag{33} \]
Where \(A\) is diaphragm area.

Bellows: an accordion‑like metal structure.

Probably the most common mechanical pressure‑to‑displacement element.
Construction:

Operation:
Most analog round‑dial pressure gauges use a Bourdon tube driving a pointer via gears.
Once we have displacement, we can convert to an electrical signal using:
Advanced differential pressure cell with feedback:

Integrated‑circuit technology enables compact solid‑state (SS) pressure sensors.

Inside:

Configurations:

Typical characteristics:
Applications:
A SS pressure sensor outputs \(25\ \mathrm{mV/kPa}\) over 0–25 kPa.
We measure the level of a liquid with density \(\rho = 1.3 \times 10^3\ \mathrm{kg/m^3}\) by mounting the sensor at the tank bottom.
Find:
Solution
At height \(h\):
\[ p = \rho g h = (1.3 \times 10^3)(9.8)(h) \]
For \(h = 2.0\ \mathrm{m}\):
\[ p = 1.3 \times 10^3 \cdot 9.8 \cdot 2.0 \approx 25.48\ \mathrm{kPa} \]
Sensor output:
\[ V = (25\ \mathrm{mV/kPa})(25.48\ \mathrm{kPa}) \approx 637\ \mathrm{mV} = 0.637\ \mathrm{V} \]
So, for 0–2.0 m: 0–0.637 V.
Sensitivity per cm (200 cm range):
\[ S = \frac{637\ \mathrm{mV}}{200\ \mathrm{cm}} \approx 3.19\ \mathrm{mV/cm} \]
For \(p < 1\) atm, down to about \(10^{-3}\) atm, two heat‑based methods:
For even lower pressures (\(10^{-3}\) to \(10^{-13}\) atm):

Kinematics & Vibration
Velocity from position: \[v(t) = \frac{dx(t)}{dt} \tag{16}\]
Acceleration from velocity/position: \[a(t) = \frac{dv(t)}{dt} = \frac{d^2 x(t)}{dt^2} \tag{17}\]
Velocity from acceleration (integral): \[v(t) = v(0) + \int_0^t a(\tau)\, d\tau \tag{18}\]
Position from velocity (integral): \[x(t) = x(0) + \int_0^t v(\tau)\, d\tau \tag{19}\]
Sinusoidal vibration: \[x(t) = x_0 \sin(\omega t) \tag{20}\]
Angular frequency–frequency relation: \[\omega = 2\pi f \tag{21}\]
Vibration velocity: \[v(t) = \omega x_0 \cos(\omega t) \tag{22}\]
Vibration acceleration: \[a(t) = -\omega^2 x_0 \sin(\omega t) \tag{23}\]
Peak acceleration: \[a_{\text{peak}} = \omega^2 x_0 \tag{24}\]
Accelerometer & Dynamics
Spring–mass equilibrium: \[ma = k\Delta x \tag{25}\]
Acceleration–displacement relation: \[a = \frac{k}{m}\Delta x \tag{26}\]
Natural frequency (Hz): \[f_N = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \tag{27}\]
Damped transient response: \[X_T(t) = X_0 e^{-\alpha t}\sin(2\pi f_N t) \tag{28}\]
Mass motion under vibration (simplified): \[\Delta x = -\frac{m x_0}{k}\omega^2 \sin(\omega t) \tag{29}\]
Pressure & Head
Gauge pressure: \[p_g = p_{\text{abs}} - p_{\text{at}} \tag{30}\]
Hydrostatic pressure (SI): \[p = \rho g h \tag{31}\]
Hydrostatic pressure (English, weight density): \[p = \rho_w h \tag{32}\]
Diaphragm force: \[F = (p_2 - p_1)A \tag{33}\]
Use Pyodide to convert acceleration from ft/s² to m/s² and g.
Visualize sinusoidal vibration position, velocity, and acceleration.
Use sliders to control frequency and peak displacement, then see the vibration responses.
Important
Notice how small changes in \(f\) have a big effect on peak acceleration because \(a_{\text{peak}} = \omega^2 x_0\).
Plot \(a_{\text{peak}} = \omega^2 x_0\) as a function of frequency for different \(x_0\).
Experiment with \(m\), \(k\) and see how natural frequency changes.
Adjust \(m\) and \(k\) via sliders and see \(f_N\) plus a suggested usable bandwidth.
Tip
As a rule of thumb, use an accelerometer up to roughly \(f_N / 2.5\) for accurate acceleration measurements.
Simulate average deceleration from a drop of height \(h\) with shock duration \(T_d\).
Use sliders for drop height and shock duration; plot average g as a function of \(h\).
Practice converting between Pa, kPa, psi, and atm.
Use sliders to explore \(p = \rho g h\) for different depths and fluid densities.
Simulate Example 17: level measurement using a SS pressure sensor.
Interactively see how tank level maps to pressure and sensor output voltage.
Use your experiments to answer:
Note
Capture these observations—they form the basis of lab reports and exam solutions.