Answer:
a) v = 25.54 ft/s
b) Xmax = 0.4180
c) Xmax = 0.0048 ft
Explanation:
a) determine the speed v at which the amplitude of the boat and the trailer will be a maximum;
to calculate the distance travelled
S = vt
given an expression of wavelength
l = vt
l = 2πv / ω
ω = 2πv / l
equation for counter of the road
y = Y sin ωt
= Y sin ( (2Ï€v / l)t)
next we calculate the angular frequency
ω = √ ( k/m)
=  √ ( g / Sst )
= √( 32.2 / ( 1.5/12))
= 16.05 rad/s
Now we calculate the speed v at which amplitude of the boat will be maximum
ω = 2πv / l
16.05 = (2 × π × v ) / 10
160.5 = 2Ï€v
v = 160.5 / 2Ï€
v = 25.54 ft/s
b)
we calculate the maximum amplitude using the following expression;
X/Y =  [ √( 1 + ( 2Sr)²)] / [ √(( 1  - r²)² +  ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×1)²)] / [ √(( 1  - 1²)² +  ( 2×0.05×r)²))]
Xmax / 0.0416 = 1.004987 / 0.1
Xmax 0.1 = 0.0418
Xmax = 0.0418 / 0.1
Xmax = 0.4180
c)
we convert speed of the boat from mph velocity m/s
v = 55 mph × 1.46667ft/s  / 1mph
v = 80.66 ft/s
next we calculate angular velocity
ω = 2πv / l
= 2π × 80.66 / 10
= 50.68 rad/s
next is the frequency ratio
r = 50.68 / 16.05
= 3.16
finally we find the amplitude of the boat
X/Y =  [ √( 1 + ( 2Sr)²)] / [ √(( 1  - r²)² +  ( 2Sr)²))]
Xmax / (0.5/12) = [ √( 1 + ( 2×0.05×3.16)²)] / [ √(( 1  - 3.16²)² +  ( 2×0.05×3.16)²))]
Xmax / 0.0416  = √1.0998 / √ ( 80.74 + 0.0998)
Xmax / 0.0416 = 1.0487 / 8.9910
Xmax 8.9910 = 0.0436
Xmax = 0.0436 / 8.9910
Xmax = 0.0048 ft
