1. Model the motion when there is no slipping

If the upper block $m$ does not slip on the lower block $M$, both move together as a single body of mass $M+m.$

The spring is attached to the lower block, so for the combined system the restoring force is $F=-kx,$ where $x$ is the displacement from mean position.

Hence the equation of motion is $(M+m)\,a=-kx$ so $a=-\frac{kx}{M+m}.$

Therefore, statement B is correct.

2. Time period of small oscillation

From $(M+m)\ddot x + kx =0,$ we get angular frequency $\omega=\sqrt{\frac{k}{M+m}}.$ So the time period is $T=2\pi\sqrt{\frac{M+m}{k}}.$

Thus statement A is correct.

3. Friction needed to move the upper block

The only horizontal force on the upper block $m$ is friction. Since it shares the same acceleration $a$, $f = m|a|.$ Using $a=-\frac{kx}{M+m},$ we get the friction force (with sign same as acceleration) $f = m\left| -\frac{kx}{M+m} \right| = \frac{mk|x|}{M+m}.$

So the magnitude of friction is $\boxed{\frac{mk|x|}{M+m}}.$

But statement C says the magnitude is $\frac{m\mu |x|}{M+m},$ which is dimensionally wrong (missing $k$ and incorrectly contains $\mu$). Hence C is false.

4. Condition for no slipping and maximum amplitude

Maximum required friction occurs at extreme position $x=A$, where acceleration magnitude is maximum: $a_{\max}=\frac{kA}{M+m}.$

Required friction on the top block is $f_{\text{req,max}}=m a_{\max}=\frac{mkA}{M+m}.$

For no slipping, $f_{\text{req,max}} \le \mu m g.$

So $\frac{mkA}{M+m} \le \mu m g.$ Cancelling $m$, $A \le \frac{\mu (M+m)g}{k}.$

Thus the maximum amplitude is $A_{\max}=\frac{\mu (M+m)g}{k}.$ So statement D is correct.

5. Maximum frictional force

The friction between the two blocks is static friction, whose maximum possible value is $f_{\max}=\mu N.$

Since the upper block of mass $m$ is resting on the lower block, the normal reaction is $N=mg.$ Thus $f_{\max}=\mu mg.$

Statement E says maximum friction can be $\mu(M+m)g,$ which is incorrect.

Hence E is false.

6. Final evaluation of statements

• A: Correct
• B: Correct
• C: False
• D: Correct
• E: False

Therefore the correct set is: $\boxed{A, B, D\text{ only}}$ which corresponds to Option C.