An Oscillator's Amplitude from Its Start

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. If the initial (t = 0) position of the particle is 1 cm and its initial velocity is ω cm/s, what is its amplitude, in cm? expects 1.41 cm

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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?

The playground

Inputs

position

x_{1}

1 cm

Outputs

amplitude

A

The solution

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 Ruby

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

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

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

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)