2. What is the period of oscillation T of the scale?
Scenario: For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.100 kg of turkey. The slices of turkey are weighed on a plate of mass 0.400 kg placed atop a vertical spring of negligible mass and force constant of 200 N/m . The slices of turkey are dropped on the plate all at the same time from a height of 0.250 m . They make a totally inelastic collision with the plate and set the scale into vertical simple harmonic motion (SHM). You may assume that the collision time is extremely small.
Potential energy: Ep = mgh
Spring energy: Esp = ½kx²
Kinetic energy: Ek = ½mv²
1. The energy that the turkey slices possess when they land on the
plate decides how much the spring compress as an amplitude while
the potential energy converts to spring energy.
The mass of the plate can be ignored since placed on the spring, it’s
already in equilibrium with the spring and does not affect the oscilla-
tions of the turkey slices.
From energy conservation: mgh = ½kA²
A = √(2mgh/k) = √ [ (2•0.1kg•9.8m/s²•0.25m)/(200N/m) ] = 4.95 cm
2. Period: T = 2π/ω
ω = v/A
The kinetic energy of the turkey slices right before hitting the plate,
which was converted from the potential energy, sets the max. speed
of the oscillation:
mgh = ½mv²
v = √(2gh)
=> ω = √(2gh)/A = √(2•9.8m/s²•0.25m)/0.0495m = 44.7 rad/s
T = 2π/44.7 = 0.14 s
Also the spring energy converts to kinetic energy of the turkey slices
during the oscillations from which it’s possible to find ω:
½kA² = ½mv²
v = √(kA²/m) = A√(k/m)
v/A = √(k/m), therefore: ω = √(k/m)
T = 2π/ω = 2π√(m/k) = 2π√(0.1kg/200N/m) = 0.14 s