A Mass on a Spring on a Table

solved SpringOscillator

Q. A spring having with a spring constant 1200 N/m $k = 1200$ N/m is mounted on a horizontal table. A mass of 3 kg $m = 3$ kg is attached to the free end of the spring. The mass is then pulled sideways to a distance of 2.0 cm $A = 2.0$ cm and released. Determine the frequency of oscillations, in hertz?

Inputs

constant

k

1200 N/m

mass

m

3 kg

distance

A

2 cm

Outputs

frequency

ν

How it's solved

Solved with SpringOscillator

Given

$k = 1200$ N/m
$m = 3$ kg
$A = 2.0$ cm

Find $ν$ .

Working

$A = 2.0$ cm $= 0.02$ m
angular frequency of a mass on a spring(block) $ω^{2} = \frac{k}{m}$
$ω^{2} = \frac{1200}{3}$
$ω = 20$ rad/s
angular frequency is a turn each oscillation(block) $ω = 2 \cdot π \cdot ν$
$20 = 2 \cdot 3.14159265 \cdot ν$
$ν = 3.183$ Hz
$ν \approx 3.2$ Hzto 2 significant figures

Conclusion

The spring's stiffness against the mass sets the angular frequency, and how far the mass is pulled out makes no difference to how often it swings.

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_mass_on_a_spring_on_a_table 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. %{frequency}, in hertz?"8  given constant: "1200 N/m"9  given mass: "3 kg"10  given distance: "2.0 cm"11  asks frequency: "Determine the frequency of oscillations"12end

solution.rb

1require_relative "../questions/a_mass_on_a_spring_on_a_table"2require_relative "../science/oscillations/spring_oscillator"34solve :a_mass_on_a_spring_on_a_table, using: SpringOscillator do5  objects :block67  take constant: block.spring_constant8  take mass: block.mass9  take distance: block.amplitude1011  frequency = block.frequency12  answer frequency:, in: "Hz", figures: 21314  conclusion "The spring's stiffness against the mass sets the angular frequency, and how far the mass is pulled " \15             "out makes no difference to how often it swings."16end

playground.rb

1require_relative "../solutions/a_mass_on_a_spring_on_a_table"23play_with :a_mass_on_a_spring_on_a_table do4  title "A Mass on a Spring on a Table"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	SimpleHarmonicMotion
position	$x$	m	—	Velocity
displacement	$Δ x$	m	—	Velocity
path length	$d$	m	—	Velocity
time taken	$Δ t$	s	0 … 1e+05	Velocity
average velocity	$v_{av}$	m/s	—	Velocity
average speed	$c_{av}$	m/s	—	Velocity
starting position	$x_{0}$	m	—	Velocity
final position	$x$	m	—	Velocity
velocity	$v$	m/s	—	Acceleration
change of velocity	$Δ v$	m/s	—	Acceleration
average acceleration	$a_{av}$	m/s²	—	Acceleration
starting velocity	$v_{0}$	m/s	—	Acceleration
final velocity	$v$	m/s	—	Acceleration
amplitude	$A$	m	0 … ∞	SimpleHarmonicMotion
stroke	$s$	m	—	SimpleHarmonicMotion
angular frequency	$ω$	rad/s	0 … ∞	SimpleHarmonicMotion
frequency	$ν$	Hz	0 … ∞	SimpleHarmonicMotion
phase constant	$φ$	—	—	SimpleHarmonicMotion
maximum speed	$v_{\max}$	m/s	—	SimpleHarmonicMotion
maximum acceleration	$a_{\max}$	m/s²	—	SimpleHarmonicMotion
acceleration	$a$	m/s²	—	SimpleHarmonicMotion
mass	$m$	kg	0 … ∞	SpringOscillator
spring constant	$k$	N/m	—	SpringOscillator
kinetic energy	$K$	J	—	SpringOscillator
potential energy	$U$	J	—	SpringOscillator
total energy	$E$	J	—	SpringOscillator
stroke of ground	$s_{g}$	m	—
amplitude of ground	$A_{g}$	m	0 … ∞
stroke of block	$s$	m	—
amplitude of block	$A$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	0 … ∞
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of block	$ω$	rad/s	0 … ∞
frequency of block	$ν$	Hz	0 … ∞
maximum speed of ground	$v_{max,g}$	m/s	—
maximum speed of block	$v_{\max}$	m/s	—
maximum acceleration of ground	$a_{max,g}$	m/s²	—
maximum acceleration of block	$a_{\max}$	m/s²	—
spring constant of ground	$k_{g}$	N/m	—
mass of ground	$m_{g}$	kg	0 … ∞
spring constant of block	$k$	N/m	—
mass of block	$m$	kg	0 … ∞
total energy of ground	$E_{g}$	J	—
total energy of block	$E$	J	—
phase constant of block	$φ$	—	—

x_{0} \equiv x_{i} (start of motion)

starting position

x \equiv x_{i} (end of motion)

final position

v_{0} \equiv v_{i} (start of motion)

starting velocity

v \equiv v_{i} (end of motion)

final velocity

d_{i} (whole) = d_{i} (part 1) + d_{i} (part 2)

path in parts

Δ t_{w} = Δ t_{p} + Δ t_{p}

time in parts

s_{b} = 2 \cdot A_{b}

stroke is twice the amplitude

ω_{b} = 2 \cdot π \cdot ν_{b}

angular frequency is a turn each oscillation

v_{max,b} = ω_{b} \cdot A_{b}

fastest through the middle

a_{max,b} = {ω_{b}}^{2} \cdot A_{b}

hardest pulled at the ends

{ω_{b}}^{2} = \frac{k_{b}}{m_{b}}

angular frequency of a mass on a spring

E_{b} = \frac{k_{b} \cdot {A_{b}}^{2}}{2}

energy of an oscillator

Δ x = x - x_{0}

change of position

v_{av} = \frac{Δ x}{Δ t}

displacement a time

c_{av} = \frac{d}{Δ t}

path a time

Δ v = v - v_{0}

velocity gained

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time

s_{g} = 2 \cdot A_{g}

stroke is twice the amplitude(ground)

s = 2 \cdot A

stroke is twice the amplitude(block)

ω_{g} = 2 \cdot π \cdot ν_{g}

angular frequency is a turn each oscillation(ground)

ω = 2 \cdot π \cdot ν

angular frequency is a turn each oscillation(block)

v_{max,g} = ω_{g} \cdot A_{g}

fastest through the middle(ground)

v_{\max} = ω \cdot A

fastest through the middle(block)

a_{max,g} = {ω_{g}}^{2} \cdot A_{g}

hardest pulled at the ends(ground)

a_{\max} = ω^{2} \cdot A

hardest pulled at the ends(block)

{ω_{g}}^{2} = \frac{k_{g}}{m_{g}}

angular frequency of a mass on a spring(ground)

ω^{2} = \frac{k}{m}

angular frequency of a mass on a spring(block)

E_{g} = \frac{k_{g} \cdot {A_{g}}^{2}}{2}

energy of an oscillator(ground)

E = \frac{k \cdot A^{2}}{2}

energy of an oscillator(block)