To find the maximum velocity of the particle, we need to find the amplitude of the simple harmonic motion (SHM).

The velocity of a particle executing SHM can be given by the equation:

v = Aωcos(ωt + φ)

Where v is the velocity, A is the amplitude, ω is the angular frequency, t is the time, and φ is the phase constant.

In this case, we are given the velocities at three points that are one foot apart. Let's assume the three points are at t=0, t=1, and t=2 (since they are one foot apart, we can assume the time difference between them is 1).

Using the equation above, we can set up three equations:

8 = Aωcos(0 + φ)
7 = Aωcos(ω + φ)
4 = Aωcos(2ω + φ)

Dividing the second equation by the first equation, we get:

7/8 = cos(ω + φ)/cos(0 + φ)

Similarly, dividing the third equation by the second equation, we get:

4/7 = cos(2ω + φ)/cos(ω + φ)

Now, we can solve these two equations simultaneously to find the values of ω and φ.

Simplifying the first equation, we get:

7cos(0 + φ) = 8cos(ω + φ)

Since cos(0) = 1, we have:

7 = 8cos(ω + φ)

Dividing the second equation by the first equation, we get:

(4/7) / (7/8) = cos(2ω + φ) / cos(ω + φ)

Simplifying, we have:

32/49 = cos(2ω + φ) / cos(ω + φ)

Now, we can use the double-angle formula for cosines to simplify the equation further:

32/49 = (2cos^2(ω + φ) - 1) / cos(ω + φ)

Multiplying both sides by cos(ω + φ), we get:

32/49 * cos(ω + φ) = 2cos^2(ω + φ) - 1

Rearranging the equation, we have:

2cos^2(ω + φ) - (32/49) * cos(ω + φ) - 1 = 0

Now, we can solve this quadratic equation for cos(ω + φ) using the quadratic formula:

cos(ω + φ) = [(32/49) ± sqrt((32/49)^2 - 4*2*(-1)) ] / (2*2)

cos(ω + φ) = [(32/49) ± sqrt(1024/2401 + 8/2)] / 4

cos(ω + φ) = [(32/49) ± sqrt(1024/2401 + 16/4)] / 4

cos(ω + φ) = [(32/49) ± sqrt(1024/2401 + 64/16)] / 4

cos(ω + φ) = [(32/49) ± sqrt(1024/2401 + 64/16)] / 4

cos(ω + φ) = [(32/49) ± sqrt(1024/2401 + 256/2401)] / 4