First at Half the Amplitude

solved SimpleHarmonicMotion · PeriodicMotion

Q. A particle moves in simple harmonic motion with amplitude 10 cm $A = 10$ cm and period 2 s $T = 2$ s. It is released from rest at its positive extreme $φ = 0$ rad. How long after release is it first 5 cm $x_{1} = 5$ cm from the centre, in seconds?

Inputs

amplitude

A

10 cm

period

T

2.00 s

position

x_{1}

5 cm

Outputs

time

Δ t

How it's solved

Solved with SimpleHarmonicMotion, PeriodicMotion

Given

$A = 10$ cm
$T = 2$ s
$x_{1} = 5$ cm
$φ = 0$ rad from “released from rest at its positive extreme”
$Δ t$ : the smallest of the values that fit from “after release is it first”

Find $Δ t$ .

Working

$A = 10$ cm $= 0.1$ m
$x_{1} = 5$ cm $= 0.05$ m
period is the time of one(particle) $T = \frac{1}{ν}$
$2 = \frac{1}{ν}$
$ν = 0.5$ Hz
angular frequency is a turn each oscillation(particle) $ω = 2 \cdot π \cdot ν$
$ω = 2 \cdot 3.14159265 \cdot 0.5$
$ω = 3.14159265$ rad/s
displacement in shm(particle, swinging) $x_{1} = A \cdot \cos (ω \cdot Δ t + φ)$
$0.05 = 0.1 \cdot \cos (3.14159265 \cdot Δ t + 0)$
$Δ t = 0.333333333, 1.66666667, 2.33333333, 3.66666667, \dots$ sthe smallest is taken
$Δ t = 0.33 s$
$Δ t \approx 0.333$ sto 3 significant figures

Conclusion

The particle passes the same point twice in every period, so many times fit; the first is a sixth of a period after release, since the cosine falls to a half at a sixth of a turn.

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 :first_at_half_the_amplitude do4  kind :numerical5  describes "A particle moves in simple harmonic motion with amplitude %{amplitude} and period %{period}. It is " \6            "released from rest at its positive extreme. %{time} %{position} from the centre, in seconds?"7  given amplitude: "10 cm"8  given period: "2 s"9  given position: "5 cm"10  asks time: "How long after release is it first"11end

solution.rb

1require_relative "../questions/first_at_half_the_amplitude"2require_relative "../science/oscillations/simple_harmonic_motion"3require_relative "../science/oscillations/periodic_motion"45solve :first_at_half_the_amplitude, using: [ SimpleHarmonicMotion, PeriodicMotion ] do6  objects :particle7  particle.goes(:swinging)89  take amplitude: particle.amplitude, period: particle.period, position: particle.final_position10  deduce_from "released from rest at its positive extreme", particle.phase_constant => "0 rad"11  deduce_from "after release is it first", swinging.time_taken => :smallest12  take time: swinging.time_taken13  answer find_value(:time, in: "s", figures: 3)1415  conclusion "The particle passes the same point twice in every period, so many times fit; the first is a sixth " \16             "of a period after release, since the cosine falls to a half at a sixth of a turn."17end

playground.rb

1require_relative "../solutions/first_at_half_the_amplitude"23play_with :first_at_half_the_amplitude do4  title "First at Half the Amplitude"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
repetitions	$N$	—	—	PeriodicMotion
period	$T$	s	—	PeriodicMotion
displacement of particle during swinging	$Δ x$	m	—
position of particle at end of swinging	$x_{1}$	m	—
position of particle at start of swinging	$x_{0}$	m	—
average velocity of particle during swinging	$v_{av}$	m/s	—
time taken of swinging	$Δ t$	s	0 … 1e+05
average speed of particle during swinging	$c_{av}$	m/s	—
path length of particle during swinging	$d$	m	—
change of velocity of particle during swinging	$Δ v$	m/s	—
velocity of particle at end of swinging from ground	$v_{1}$	m/s	—
velocity of particle at start of swinging from ground	$v_{0}$	m/s	—
average acceleration of particle during swinging	$a_{av}$	m/s²	—
frequency of particle	$ν$	Hz	0 … ∞
repetitions of particle during swinging	$N$	—	—
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 … ∞
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	—
phase constant of particle	$φ$	—	—
acceleration of particle at end of swinging	$a_{1}$	m/s²	—
period of ground	$T_{g}$	s	—
period of particle	$T$	s	—

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

T_{b} = \frac{1}{ν_{b}}

period is the time of one

Δ x = x_{1} - x_{0}

change of position(swinging)

v_{av} = \frac{Δ x}{Δ t}

displacement a time(swinging)

c_{av} = \frac{d}{Δ t}

path a time(swinging)

Δ v = v_{1} - v_{0}

velocity gained(swinging)

a_{av} = \frac{Δ v}{Δ t}

velocity gained a time(swinging)

ν = \frac{N}{Δ t}

frequency is repetitions a second(swinging)

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, swinging)

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

velocity in shm(particle, swinging)

a_{1} = - ω^{2} \cdot x_{1}

acceleration in shm(particle, swinging)

T_{g} = \frac{1}{ν_{g}}

period is the time of one(ground)

T = \frac{1}{ν}

period is the time of one(particle)