I'm not a $D$-module person. I'm hoping someone else can give a slightly more insightful explanation. (It looks like YBL gives a clear picture of the holonomic case.)
By definition $Ch(M)$ is the support of the associated graded of $M$, for a suitable filtration, so it should lie in the preimage of the support of $M$. However, in many interesting cases the inclusion would be strict. If $M$ is holonomic, $Ch(M)$ is Lagrangian, so it wouldn't coincide with the preimage of $supp(M)$. The simplest case where this
happens is when $M$ is a flat connection, then $Ch(M)$ is the zero section of $T^*X$.
Continuation: Perhaps it's worthwhile making this a little more explicit.
The simplest example is $X=mathbb{A}^n$. Then (the global sections of) $D_X$
is the Weyl algebra with generators $x_1,ldots, x_n, partial_1,ldots, partial_n$ with
commutation relations $[x_i,partial_j]= delta_{ij}$.
If we force these to commute, by passing to the associated graded with respect to
the filtration by order of operators, we obtain the polynomial ring in $2n$ variables
or in other words the coordinate ring of $T^*mathbb{A}^n$. Any finitely generated $D_X$-module $M$, carries a (noncanonical) compatible filtration, so can define $Ch(M)= Supp(GrM)subset T^*mathbb{A}^n$ (it is independent of the filtration). Now let
$M= D/sum Dpartial_i$, which corresponds to $mathcal{O}_X$. I'll omit the
details, but one can see that
$Ch(M) = V(partial_1,ldots partial_n)$ (the zero section), and $pi^*supp(M) = T^*mathbb{A}^n$ where
$pi:T^*mathbb{A}^nto mathbb{A}^n$.
For a nonholonomic example, take $M= D_X$. Then $Ch(M)= pi^{-1}supp(M) = T^*mathbb{A}^n$.
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