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 0.60 s . His older sister pulls the spring a bit more than intended. She pulls the animal 27 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

0.407 m

Step-by-step explanation:

T = Time period = 0.6 s

A = Amplitude = 27 cm

m = Mass = 120 g

Angular speed is given by

`\omega=(2\pi)/(T)\\\Rightarrow \omega=(2\pi)/(0.6)\\\Rightarrow \omega=10.47\ rad/s`

Spring constant is given by

`k=m\omega^2\\\Rightarrow k=0.12* 10.47^2\\\Rightarrow k=13.15\ N/m`

As the energy of the system is conserved we have

`(1)/(2)kA^2=mgh\\\Rightarrow h=(1)/(2mg)kA^2\\\Rightarrow h=(1)/(2* 0.12* 9.81)* 13.15* 0.27^2\\\Rightarrow h=0.407\ m`

The animal will go 0.407 m above its equilibrium position.