Question

In a control system, an accelerometer consists of a 4.7 g object sliding on a calibrated horizontal rail. A low-mass spring attaches the object to a flange at one end of the rail. Grease on the rail makes static friction negligible, but rapidly damps out vibrations of the sliding object. When subject to a steady acceleration of

, the object should be at a location away from its equilibrium position. Find the force constant of the spring required for the calibration to be correct.

Short answer

The force constant of the spring required for the calibration to be correct is 7.37 N/M

Step-by-step solution

Given data

Mass of the object.

$$\begin{gathered} m=4.7\text{g} \\ =4.7·\left(1\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right) \\ =0.0047\text{kg} \end{gathered}$$

Displacement of the object

$$\begin{gathered} x=0.5\text{cm} \\ =0.5·\left(1\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right) \\ =0.005\text{m} \end{gathered}$$

Required acceleration at this displacement

$$\begin{gathered} a=0.8g \\ =0.8\times 9.8\frac{\text{m}}{{\text{s}}^2} \\ =7.84\frac{\text{m}}{{\text{s}}^2} \end{gathered}$$

Step 2:Spring force and second law of motion

The force that tries to bring a spring to its equilibrium position when it is extended by distance is

F=-Kx

Here, K is the spring constant.

From second law of motion, the applied force on a body of mass m and its acceleration a are related as

F=ma

Step 3:Determinethe spring constant

Eliminate the force from equations (I) and (II) to get

$$\begin{gathered} kx=ma \\ k=\frac{ma}{x} \end{gathered}$$

Substitute the values to get

$$\begin{gathered} k=\frac{0.0047\text{kg}\times 7.84\frac{\text{m}}{{\text{s}}^2}}{0.005\text{m}} \\ =7.37·\left(1\frac{\text{kg}·{\text{m}}^2}{{\text{s}}^2}\times \frac{1\text{N}}{1\frac{\text{kg}·{\text{m}}^2}{{\text{s}}^2}}\right)·\left(\frac{1}{1\text{m}}\right) \\ =7.37\frac{\text{N}}{\text{m}} \end{gathered}$$

Thus the spring constant is 7.37 N/m