First at Half the Amplitude

Q. A particle moves in simple harmonic motion with amplitude 10 cm and period 2 s. It is released from rest at its positive extreme. How long after release is it first 5 cm from the centre, in seconds? expects 0.333 s

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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?

The playground

Inputs

amplitude

A

10 cm

period

T

2.00 s

position

x_{1}

5 cm

Outputs

time

Δ t

The solution

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 Ruby

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

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

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

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)