Question

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 0.54 s . His older sister pulls the spring a bit more than intended. She pulls the animal 32 cm below its equilibrium position, then lets go. The animal flies upward and detaches from the spring right at the animal's equilibrium position. Part A If the animal does not hit anything on the way up, how far above its equilibrium position will it go

Answer 1

the wooden animal will go 0.7068 m above its equilibrium

Step-by-step explanation:

Given the data in the question;

mass of wooden animal m = 120 g = 0.12 kg

the animal oscillates up and down, T = 0.54 s

older sister pulls the animal 32 cm below its equilibrium position;

x = 32 cm = 0.32 m

g = 9.81 m/s²

We know that

k = mω²

where ω = 2π/T

So, k = m( 2π/T )²

we substitute

k = 0.12( 2π / 0.54 )²

k = 0.12 × (11.6355)²

k = 0.12 × 135.38486

k = 16.25 N/c

so Also,

kx²/2 = mgh

we solve for h

h = kx² / 2mg

we substitute

h = ( 16.25 × (0.32)²) / ( 2 × 0.12 × 9.81 )

h = 1.664 / 2.3544

h = 0.7068 m

Therefore, the wooden animal will go 0.7068 m above its equilibrium