smaller-volume-change-lowers-pressure - Elasticity and Nonlinear Dynamics Step by Step

Because Delta-P scales with Delta-V / V at fixed K, a smaller volume change produces a smaller pressure change. The trend is exact only for the stated bulk modulus, volume change, and original volume in the ideal linear elastic model. Real materials add nonlinear response, porosity, temperature/loading history, geometry tolerance, calibration, and measurement/source uncertainty.

For the same bulk modulus and volume, pressure change follows volume change. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Keep bulk modulus and starting volume fixed

This contrast changes only the volume-change marker. The smaller marker should produce a smaller pressure change.

K=12\ \text{Pa}\qquad \Delta V=\tfrac{1}{2}\ \text{m}^{3}

Smaller volume changes make smaller pressure changes

The table keeps the same starting volume and modulus. Each row is an exact positive-magnitude compression ledger.

\begin{array}{c|c|c|c}V&\Delta V&K&\Delta P\\6\ \text{m}^{3}&\tfrac{1}{4}\ \text{m}^{3}&12\ \text{Pa}&\tfrac{1}{2}\ \text{Pa}\\6\ \text{m}^{3}&\tfrac{1}{2}\ \text{m}^{3}&12\ \text{Pa}&1\ \text{Pa}\\6\ \text{m}^{3}&1\ \text{m}^{3}&12\ \text{Pa}&2\ \text{Pa}\\\end{array}

A smaller volume change lowers the pressure change

The modulus and starting volume stay fixed; only the checked volume change is smaller.

\Delta P=12\ \text{Pa}\cdot\frac{\tfrac{1}{2}\ \text{m}^{3}}{6\ \text{m}^{3}}=1\ \text{Pa}