In an engine the piston undergoes vertical simple harmonic motion with amplitude 7 cm. A washer rests on top of the piston and moves with it. The motor speed is slowly increased. The frequency of the piston at which the washer no longer stays in contact with the piston, is close to:

We begin by recalling that the vertical motion of the piston is simple harmonic. For a particle in simple harmonic motion we write its displacement from the mean position as

$$y = A \sin(\omega t),$$

where $$A$$ is the amplitude of oscillation and $$\omega$$ is the angular frequency. The acceleration is the second time-derivative of the displacement. Differentiating twice, we have the well-known SHM result

$$a = -\omega^2\,y.$$

The magnitude of the greatest possible acceleration is obtained when $$|y| = A$$, giving

$$a_{\text{max}} = \omega^2 A.$$

Now we concentrate on the washer resting on the piston. The washer will remain in contact as long as the normal reaction between the washer and the piston is positive. The normal reaction becomes zero precisely when the piston accelerates downward with an acceleration equal to the acceleration due to gravity $$g$$. At that instant, the washer is on the verge of leaving the piston. Therefore the critical condition for losing contact is

$$a_{\text{max}} = g.$$

Substituting the SHM expression for the maximum acceleration, we get

$$\omega^2 A = g.$$

We know the amplitude $$A$$ is 7 cm, so we first convert it into metres:

$$A = 7 \text{ cm} = 0.07 \text{ m}.$$

Putting the numerical values into the condition, we have

$$\omega^2 \times 0.07 = 9.8.$$

Solving for $$\omega^2$$,

$$\omega^2 = \frac{9.8}{0.07} = 140.$$ $$\Rightarrow \omega = \sqrt{140}\ \text{rad s}^{-1}.$$

But the frequency $$f$$ is related to the angular frequency by the formula

$$f = \frac{\omega}{2\pi}.$$

Substituting $$\omega = \sqrt{140}$$, we get

$$f = \frac{\sqrt{140}}{2\pi}.$$ To evaluate this, we write

$$\sqrt{140} \approx 11.832,$$

and since $$2\pi \approx 6.283$$,

$$f \approx \frac{11.832}{6.283} \approx 1.884\ \text{Hz}.$$

This numerical value rounds to about 1.9 Hz, which matches Option B in the list provided.

Hence, the correct answer is Option B.