Answer
$$2.00$$
Work Step by Step
When $\phi=0$ it is obvoius that superposition wave has amplitude $2 \Delta p_{m}$ .
For other cases, it is useful to write
$$
\Delta p_{1}+\Delta p_{2}=\Delta p_{m}(\sin (\omega t)+\sin (\omega t-\phi))=\left(2 \Delta p_{m} \cos \frac{\phi}{2}\right) \sin \left(\omega t-\frac{\phi}{2}\right)
$$
The factor in front of sine function gives amplitude $\Delta p_{r}$ . Therefore,
$$\Delta p_{r} / \Delta p_{m}=2 \cos (\phi / 2) .$$
When $\phi=0,$
$$ \Delta p_{r} / \Delta p_{m}=2 \cos (0)=2.00$$
