1. Choose the direction of motion

Let the block move along the smooth incline. Since the plane is smooth, the motion is only along the incline.

2. Understand the equilibrium position

The component of gravity along the incline is $Mg\sin\alpha$ which shifts the equilibrium position of the block.

However, frequency of SHM depends on the restoring force about equilibrium, not on the constant force itself.

3. Restoring force when displaced

Suppose the block is displaced by a small distance $x$ along the incline from its equilibrium position.

• The spring on one side gets compressed/stretched by $x$.
• The spring on the other side also changes length by $x$ in such a way that it exerts a restoring force toward equilibrium.

Each spring contributes a restoring force of magnitude $kx$ Therefore, total restoring force is $F = -kx-kx = -2kx$

So the effective spring constant is $k_{\text{eff}}=2k$

4. Equation of motion

Hence, $M\ddot x = -2kx$

or $\ddot x + \frac{2k}{M}x=0$

Comparing with standard SHM equation $\ddot x + \omega^2 x=0$ we get $\omega = \sqrt{\frac{2k}{M}}$

5. Frequency

Therefore, $f=\frac{\omega}{2\pi}=\frac{1}{2\pi}\sqrt{\frac{2k}{M}}$

6. Match with options

This corresponds to:

Option C: $\frac{1}{2\pi}\sqrt{\frac{2k}{M}}$

7. Comparison with stored answer

Derived answer = C

Stored correct answer = C

They agree.