At what frequency does the driver of the car hear the ambulance’s siren? the velocity

a) r eq = -a/(2b)

b) k = a/r eq = -2b

Explanation:

since

U(r) = ar + br²

a) the equilibrium position dU/dr = 0

U(r) = a + 2br = 0 →  r eq= -a/2b

b) the Taylor expansion around the equilibrium position is

U(r) = U(r eq) + ∑ Un(r eq) (r- r eq)^n / n!  

,where Un(a) is the nth derivative of U respect with r , evaluated in a

Since the 3rd and higher order derivatives are =0 , we can expand until the second derivative

U(r) = U(r eq) + dU/dr(r eq) (r- r eq) +  d²U/dr²(r eq) (r- r eq)² /2

since dU/dr(r eq)=0

U(r) = U(r eq) + d²U/dr²(r eq) (r- r eq)² /2

comparing with an energy balance of a spring around its equilibrium position

U(r) - U(r eq)  = 1/2 k (r-r eq)² → U(r) = U(r eq)  + 1/2 k (r-r eq)²

therefore we can conclude

k = d²U/dr²(r eq) = -2b , and since r eq = -a/2b → -2b=a/r eq

thus

k= a/r eq