AC DC Analysis of diode
DC analysis:

The voltage drop across the diode is 0.7 volts, and the DC current through the diode is only determined by the resistor:

$$IDQ=\frac{7V-0.7V}{4700\Omega}=1.34mA$$

For the AC analysis you can use the differential resistance

$$r_d=\frac{kT}{qIDQ}\approx\frac{0.025}{IDQ}\Omega=18.65\Omega\quad\text{(at room temperature)}$$

So for \$v_D(t)\$ we get

$$v_D(t)=-\frac{r_d}{r_d+4700\Omega}0.4\sin(\omega t)V=-1.58\sin(\omega t)mV$$