Question

III INT An infant's toy has a 120 g wooden animal hanging from a spring. If pulled down gently, the animal oscillates up and down with a period of 0.50 s. His older sister pulls the spring a bit more than intended. She pulls the animal 30 cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. If the animal does not hit anything on the way up, how far above its equilibrium position will it go?

Answer 1

To find the maximum displacement of the wooden animal from its equilibrium position in the infant's toy, calculate the amplitude of the oscillation.

Step-by-step explanation:

In this problem, we are given the period of oscillation (T) of an infant's toy and the mass of the wooden animal hanging from the spring. The period of oscillation can be used to calculate the angular frequency (ω) of the oscillation using the equation ω = 2π/T. Once we have the angular frequency, we can use the equation for the position of an object undergoing simple harmonic motion to find the maximum displacement from its equilibrium position. The maximum displacement, or amplitude (A), is given by A = xmax - xeq, where xeq is the equilibrium position of the spring and xmax is the maximum displacement from the equilibrium position.

Given that the period of oscillation is 0.50 s, the angular frequency is ω = 2π / 0.50 = 4π rad/s. The maximum displacement can be calculated using the equation A = xmax - xeq. Since the animal detaches from the spring right at its equilibrium position, the maximum displacement is equal to 30 cm. So, the amplitude of the oscillation is 30 cm.