Answer:
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
