Solution:

For the SHM given by:

\[
y = A \cos(\omega t + \frac{\pi}{4})
\]

Solution:

1. Velocity: The velocity \( v \) is the derivative of \( y \) with respect to \( t \):
\[
v = \frac{dy}{dt} = -A \omega \sin(\omega t + \frac{\pi}{4})
\]

2. Maximum Speed: The speed will be maximum when \( \sin(\omega t + \frac{\pi}{4}) = \pm 1 \).

Therefore,
\[
\omega t + \frac{\pi}{4} = \frac{\pi}{2}
\]

3. Solving for \( t \):
\[
\omega t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}
\]

\[
t = \frac{\pi}{4\omega}
\]

Answer:
The speed will be maximum at \( t = \frac{\pi}{4\omega} \).