An Oscillator's Amplitude from Its Start

solved SimpleHarmonicMotion

Q. The motion of a particle executing simple harmonic motion is described by the displacement function, x(t) = A cos (ωt + φ). The angular frequency of the particle is π s–1 $ω = 0.5$ rev/s. If the initial (t = 0) $Δ t = 0$ s position of the particle is 1 cm $x_{1} = 1$ cm and its initial velocity is ω cm/s $v_{1} = 0.5$ rev cm/s, what is its amplitude, in cm?

Inputs

position

x_{1}

1 cm

Outputs

amplitude

A

How it's solved

Solved with SimpleHarmonicMotion

Given

$x_{1} = 1$ cm
$Δ t = 0$ s from “initial (t = 0)”
$ω = 0.5$ rev/s from “π s–1”
$v_{1} = 0.5$ rev cm/s from “ω cm/s”

Find $A$ .

Working

$x_{1} = 1$ cm $= 0.01$ m
$ω = 0.5$ rev/s $= 3.14159265$ rad/s
$v_{1} = 0.5$ rev cm/s $= 0.0314159265$ m/s
amplitude from where and how fast(particle, oscillating) $A^{2} = {x_{1}}^{2} + {(\frac{v_{1}}{ω})}^{2}$
$A^{2} = {0.01}^{2} + {(\frac{0.0314159265}{3.14159265})}^{2}$
$A = 1.414$ cm
$A = 0.0141421356$ m $= 1.41421356$ cm
$A \approx 1.41$ cmto 3 significant figures

Conclusion

At the start the position and the velocity over the angular frequency are the two sides of a right angle, and the amplitude is its hypotenuse.

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 :an_oscillator_s_amplitude_from_its_start do4  kind :numerical5  describes "The motion of a particle executing simple harmonic motion is described by the displacement function, " \6            "x(t) = A cos (ωt + φ). The angular frequency of the particle is π s–1. If the initial (t = 0) position " \7            "of the particle is %{position} and its initial velocity is ω cm/s, %{amplitude}, in cm?"8  given position: "1 cm"9  asks amplitude: "what is its amplitude"10end

solution.rb

1require_relative "../questions/an_oscillator_s_amplitude_from_its_start"2require_relative "../science/oscillations/simple_harmonic_motion"34solve :an_oscillator_s_amplitude_from_its_start, using: SimpleHarmonicMotion do5  objects :particle6  axes :x7  particle.goes(:oscillating)89  take position: particle.final_position1011  deduce_from "initial (t = 0)", oscillating.time_taken => "0 s"12  deduce_from "π s–1", particle.angular_frequency => "0.5 rev/s"13  deduce_from "ω cm/s", particle.final_velocity => "0.5 rev cm/s"1415  amplitude = particle.amplitude16  answer amplitude:, in: "cm", figures: 31718  conclusion "At the start the position and the velocity over the angular frequency are the two sides of a " \19             "right angle, and the amplitude is its hypotenuse."20end

playground.rb

1require_relative "../solutions/an_oscillator_s_amplitude_from_its_start"23play_with :an_oscillator_s_amplitude_from_its_start do4  title "An Oscillator's Amplitude from Its Start"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
displacement of particle during oscillating along x	$Δ x$	m	—
position of particle at end of oscillating along x	$x_{1}$	m	—
position of particle at start of oscillating along x	$x_{0}$	m	—
average velocity of particle during oscillating along x	$v_{av}$	m/s	—
time taken of oscillating	$Δ t$	s	0 … 1e+05
average speed of particle during oscillating	$c_{av}$	m/s	—
path length of particle during oscillating	$d$	m	—
change of velocity of particle during oscillating along x	$Δ v$	m/s	—
velocity of particle at end of oscillating from ground along x	$v_{1}$	m/s	—
velocity of particle at start of oscillating from ground along x	$v_{0}$	m/s	—
average acceleration of particle during oscillating along x	$a_{av}$	m/s²	—
stroke of ground	$s_{g}$	m	—
amplitude of ground	$A_{g}$	m	0 … ∞
stroke of particle	$s$	m	—
amplitude of particle	$A$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	0 … ∞
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of particle	$ω$	rad/s	0 … ∞
frequency of particle	$ν$	Hz	0 … ∞
maximum speed of ground	$v_{max,g}$	m/s	—
maximum speed of particle	$v_{\max}$	m/s	—
maximum acceleration of ground	$a_{max,g}$	m/s²	—
maximum acceleration of particle	$a_{\max}$	m/s²	—
position of particle at end of oscillating	$\| x_{1} \|$	m	—
phase constant of particle	$φ$	—	—
velocity of particle at end of oscillating from ground	$\| v_{1} \|$	m/s	—
acceleration of particle at end of oscillating	$\| a_{1} \|$	m/s²	—

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

Δ x = x_{1} - x_{0}

change of position(oscillating, x)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(oscillating, x)

c_{av} = \frac{d}{Δ t}

path a time(oscillating, x)

Δ v = v_{1} - v_{0}

velocity gained(oscillating, x)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(oscillating, x)

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

stroke is twice the amplitude(ground)

s = 2 \cdot A

stroke is twice the amplitude(particle)

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(particle)

v_{max,g} = ω_{g} \cdot A_{g}

fastest through the middle(ground)

v_{\max} = ω \cdot A

fastest through the middle(particle)

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(particle)

| x_{1} | = A \cdot \cos (ω \cdot Δ t + φ)

displacement in shm(particle, oscillating)

| v_{1} | = - A \cdot ω \cdot \sin (ω \cdot Δ t + φ)

velocity in shm(particle, oscillating)

| a_{1} | = - ω^{2} \cdot | x_{1} |

acceleration in shm(particle, oscillating)

A^{2} = {x_{1}}^{2} + {(\frac{v_{1}}{ω})}^{2}

amplitude from where and how fast(particle, oscillating)