The bug will cross the sticky region once in each cycle of its motion, where a cycle is defined as one complete round trip from the top of the bowl to the bottom and back to the top.
To find the number of cycles the bug goes through, we can use conservation of mechanical energy. At the top of the bowl, the bug has only potential energy, which is converted to kinetic energy as it slides down the bowl. At the bottom of the bowl, all of the potential energy has been converted to kinetic energy, and as the bug slides up the other side of the bowl, the kinetic energy is converted back into potential energy. At the top of the bowl again, the bug has only potential energy, and the cycle repeats.
Because there is no friction (except for the sticky patch), the total mechanical energy of the system is conserved. Therefore, the potential energy at the top of the bowl is equal to the potential energy at the bottom of the bowl, and the kinetic energy at the bottom of the bowl is equal to the kinetic energy at the top of the bowl.
We can set the potential energy at the top of the bowl to zero, and use the conservation of energy to find the potential energy at the bottom of the bowl:
mgh = (1/2)mv^2
where m is the mass of the bug, g is the acceleration due to gravity, h is the depth of the bowl, and v is the speed of the bug at the bottom of the bowl.
Solving for v, we get:
v = sqrt(2gh)
Plugging in the numbers, we get:
v = sqrt(29.810.12) = 0.775 m/s
The time it takes for the bug to slide from the top of the bowl to the bottom and back up to the top is twice the time it takes to slide from the top to the bottom:
t = 2sqrt(2h/g) = 2sqrt(2*0.12/9.81) = 0.774 s
Therefore, the frequency of the bug's motion is:
f = 1/t = 1/0.774 = 1.29 Hz
Since the bug completes one cycle in each oscillation, the bug will cross the sticky region 1.29 times per second, or approximately once every 0.78 seconds.