Final answer:
The ratio of the frequency of oscillation with an adult to the frequency with the child on a bungee cord is 1/√5, as the adult's mass is five times greater than the child's and frequency is inversely proportional to the square root of mass.
Step-by-step explanation:
When a child plays on a bungee cord and oscillates with a certain frequency f, and then an adult with a mass that is five times greater uses the same bungee cord, the frequency of oscillation changes due to the difference in mass.
The frequency of oscillation for a mass on a spring (or in this case, a bungee cord) is given by the formula f = 1 / (2π) × √(k/m), where k is the spring constant and m is the mass. Since the spring constant k does not change, and the adult's mass is five times that of the child, the frequency ratio would be √(1/5) of the child's frequency because frequency is inversely proportional to the square root of mass.
If the mass is increased by a factor of five, then the frequency with which the adult oscillates will be 1/√5 times the frequency with which the child oscillates. Hence, the ratio of the adult's frequency to the child's frequency is 1/√5.