Answer
The inductance in circuit 1 is less than the inductance in circuit 2.
Work Step by Step
We can write an expression for the period of oscillation:
$T = \frac{2\pi}{\omega} = 2\pi~\sqrt{LC}$
From the graph, we can see that circuit 1 has a shorter period of oscillation than circuit 2.
Therefore, the inductance $L$ in circuit 1 is less than the inductance in circuit 2.
