play phryzby with physics

A Spring's Mass at Its Hardest Pull

solved SpringOscillator

Q. A spring having with a spring constant 1200 N/mk=1200N/m is mounted on a horizontal table. A mass of 3 kgm=3kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cmA=2.0cm and released. Determine the maximum acceleration of the mass, in m/s²?

−AO+AA = 0.02 mm = 3 kgk = 1200 N/ma_max = ?
Inputs
k
1200 N/m
m
3 kg
A
2 cm
Outputs
accelerationamax?

How it's solved

Solved with SpringOscillator

Given

  • k=1200N/m
  • m=3kg
  • A=2.0cm

Find amax.

Working
  1. A=2.0cm=0.02m
  2. angular frequency of a mass on a spring(block)ω2=km
  3. ω2=12003
  4. ω=20rad/s
  5. hardest pulled at the ends(block)amax=ω2⋅A
  6. amax=202⋅0.02
  7. amax=8m/s²
  8. amax≈8.0m/s²to 2 significant figures
Conclusion

The spring pulls hardest where it is stretched most, at the ends of the swing, and there the acceleration is the angular frequency squared times the amplitude.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb
1require_relative "../physics"23question :a_spring_s_mass_at_its_hardest_pull do4  kind :numerical5  describes "A spring having with a spring constant %{constant} is mounted on a horizontal table. A mass of %{mass} " \6            "is attached to the free end of the spring. The mass is then pulled sideways to a distance of " \7            "%{distance} and released. %{acceleration}, in m/s²?"8  given constant: "1200 N/m"9  given mass: "3 kg"10  given distance: "2.0 cm"11  asks acceleration: "Determine the maximum acceleration of the mass"12end
solution.rb
1require_relative "../questions/a_spring_s_mass_at_its_hardest_pull"2require_relative "../science/oscillations/spring_oscillator"34solve :a_spring_s_mass_at_its_hardest_pull, using: SpringOscillator do5  objects :block67  take constant: block.spring_constant8  take mass: block.mass9  take distance: block.amplitude1011  acceleration = block.maximum_acceleration12  answer acceleration:, in: "m/s2", figures: 21314  conclusion "The spring pulls hardest where it is stretched most, at the ends of the swing, " \15             "and there the acceleration is the angular frequency squared times the amplitude."16end
playground.rb
1require_relative "../solutions/a_spring_s_mass_at_its_hardest_pull"23play_with :a_spring_s_mass_at_its_hardest_pull do4  title "A Spring's Mass at Its Hardest Pull"5end
The laws it drew on
quantitywrittenunitswithinfrom
piπ—0 … ∞SimpleHarmonicMotion
positionxm—Velocity
displacementΔxm—Velocity
path lengthdm—Velocity
time takenΔts0 … 1e+05Velocity
average velocityvavm/s—Velocity
average speedcavm/s—Velocity
starting positionx0m—Velocity
final positionxm—Velocity
velocityvm/s—Acceleration
change of velocityΔvm/s—Acceleration
average accelerationaavm/s²—Acceleration
starting velocityv0m/s—Acceleration
final velocityvm/s—Acceleration
amplitudeAm0 … ∞SimpleHarmonicMotion
strokesm—SimpleHarmonicMotion
angular frequencyωrad/s0 … ∞SimpleHarmonicMotion
frequencyνHz0 … ∞SimpleHarmonicMotion
phase constantφ——SimpleHarmonicMotion
maximum speedvmaxm/s—SimpleHarmonicMotion
maximum accelerationamaxm/s²—SimpleHarmonicMotion
accelerationam/s²—SimpleHarmonicMotion
massmkg0 … ∞SpringOscillator
spring constantkN/m—SpringOscillator
kinetic energyKJ—SpringOscillator
potential energyUJ—SpringOscillator
total energyEJ—SpringOscillator
stroke of groundsgm—
amplitude of groundAgm0 … ∞
stroke of blocksm—
amplitude of blockAm0 … ∞
angular frequency of groundωgrad/s0 … ∞
frequency of groundνgHz0 … ∞
angular frequency of blockωrad/s0 … ∞
frequency of blockνHz0 … ∞
maximum speed of groundvmax,gm/s—
maximum speed of blockvmaxm/s—
maximum acceleration of groundamax,gm/s²—
maximum acceleration of blockamaxm/s²—
spring constant of groundkgN/m—
mass of groundmgkg0 … ∞
spring constant of blockkN/m—
mass of blockmkg0 … ∞
total energy of groundEgJ—
total energy of blockEJ—
phase constant of blockφ——
x0≡xi(start of motion)
starting position
x≡xi(end of motion)
final position
v0≡vi(start of motion)
starting velocity
v≡vi(end of motion)
final velocity
di(whole)=di(part 1)+di(part 2)
path in parts
Δtw=Δtp+Δtp
time in parts
sb=2⋅Ab
stroke is twice the amplitude
ωb=2⋅π⋅νb
angular frequency is a turn each oscillation
vmax,b=ωb⋅Ab
fastest through the middle
amax,b=ωb2⋅Ab
hardest pulled at the ends
ωb2=kbmb
angular frequency of a mass on a spring
Eb=kb⋅Ab22
energy of an oscillator
Δx=x-x0
change of position
vav=ΔxΔt
displacement a time
cav=dΔt
path a time
Δv=v-v0
velocity gained
aav=ΔvΔt
velocity gained a time
sg=2⋅Ag
stroke is twice the amplitude(ground)
s=2⋅A
stroke is twice the amplitude(block)
ωg=2⋅π⋅νg
angular frequency is a turn each oscillation(ground)
ω=2⋅π⋅ν
angular frequency is a turn each oscillation(block)
vmax,g=ωg⋅Ag
fastest through the middle(ground)
vmax=ω⋅A
fastest through the middle(block)
amax,g=ωg2⋅Ag
hardest pulled at the ends(ground)
amax=ω2⋅A
hardest pulled at the ends(block)
ωg2=kgmg
angular frequency of a mass on a spring(ground)
ω2=km
angular frequency of a mass on a spring(block)
Eg=kg⋅Ag22
energy of an oscillator(ground)
E=k⋅A22
energy of an oscillator(block)