Measuring the electrical performance of a low pass filter
The problem arises from adding the \$7.5 \Omega\$ at the output. If you calculate the Thevenin Equivalent you'll get the following circuit, where the \$5 \Omega\$ resistor is the parallel equivalent of \$7.5 \Omega\$ and \$15.5 \Omega\$.







simulate this circuit &ndash; Schematic created using CircuitLab

You can see that the output voltage has an insertion loss from the factor \$\frac {7.5} {7.5+15.5}\$ and it is where the 9.6 dB comes from.

The \$5 \Omega\$ equivalent resistor explains why the frequency cutoff changes to 2100 Hz.