A ball of density $d$ is dropped on to a horizontal solid surface. It bounces elastically from the surface and returns to its original position in a time $t_1$. Next, the ball is released and it falls through the same height before striking the surface of a liquid of density of $d_L$ (a) If $d < d_L$, obtain an expression (in terms of $d$, $t_1$ and $d_L$) for the time $t_2$ the ball takes to come back to the position from which it was released. (b) Is the motion of the ball simple harmonic? (c) If $d = d_L$, how does the speed of the ball depend on its depth inside the liquid? Neglect all frictional and other dissipative forces. Assume the depth of the liquid to be large.

Step 1: Analyze the motion of the ball in air.

The ball is dropped from a height $h$ and returns to its original position in time $t_1$, meaning the time of fall is $\frac{t_1}{2}$.

$$v_0 = g \left(\frac{t_1}{2}\right)$$ $$t_1 = \frac{2v_0}{g}$$

The height $h$ from which the ball is dropped is:

$$h = \frac{v_0^2}{2g} = \frac{g t_1^2}{8}$$

Step 2: Determine the acceleration of the ball inside the liquid.

When the ball enters the liquid, it experiences a downward gravitational force and an upward buoyant force.

$$W = dVg$$ $$F_B = d_L Vg$$

Since $d < d_L$, the net upward force is constant and produces an upward acceleration $a$.

$$F_{\text{net}} = (d_L - d)Vg$$ $$a = \frac{F_{\text{net}}}{dV} = g\left(\frac{d_L - d}{d}\right)$$

Step 3: Calculate the time spent by the ball inside the liquid.

The ball enters the liquid with speed $v_0$ and decelerates to a stop before returning to the surface.

$$t_{\text{liq}} = \frac{2v_0}{a}$$

Substitute the expression for $a$ and use $t_1 = \frac{2v_0}{g}$.

$$t_{\text{liq}} = \frac{2v_0 d}{g(d_L - d)}$$ $$t_{\text{liq}} = \frac{d t_1}{d_L - d}$$

Step 4: Find the total return time $t_2$.

The total time $t_2$ is the sum of the time spent in the air and the time spent in the liquid.

$$t_2 = t_1 + t_{\text{liq}}$$ $$t_2 = t_1 + \frac{d t_1}{d_L - d}$$ $$t_2 = t_1 \left(\frac{d_L}{d_L - d}\right)$$

Step 5: Determine if the motion is simple harmonic.

For simple harmonic motion, the restoring force must be proportional to the displacement.

$$F_{\text{net}} = (d_L - d)Vg$$

Since the net force inside the liquid is constant and does not depend on depth, the motion is not simple harmonic.

Step 6: Analyze the speed of the ball when $d = d_L$.

If the density of the ball equals the density of the liquid, the net force acting on the ball inside the liquid is zero.

$$F_{\text{net}} = (d_L - d)Vg = 0$$

Since there is no acceleration, the ball moves at a constant speed equal to its entry speed $v_0$ at all depths.

$$v = v_0 = \frac{g t_1}{2}$$

Answer: (a) $t_2 = t_1 \left(\dfrac{d_L}{d_L - d}\right)$ (b) No, the motion is not simple harmonic. (c) The speed is constant and independent of depth, equal to $\dfrac{g t_1}{2}$.