A Locomotive Piston

Q. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min, what is its maximum speed, in m/s? expects 1.7 m/s

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Q. The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of 1.0 m $s = 1.0$ m. If the piston moves with simple harmonic motion with an angular frequency of 200 rad/min $ω = 200$ rad/min, what is its maximum speed, in m/s?

The playground

Inputs

stroke

s

1.00 m

angular frequency

ω

200 rad/min

Outputs

speed

v_{\max}

The solution

Solved with SimpleHarmonicMotion

Given

$s = 1.0$ m
$ω = 200$ rad/min

Find $v_{\max}$ .

Working

$ω = 200$ rad/min $= 3.33333333$ rad/s
stroke is twice the amplitude(piston) $s = 2 \cdot A$
$1 = 2 \cdot A$
$A = 0.5$ m
fastest through the middle(piston) $v_{\max} = ω \cdot A$
$v_{\max} = 3.33333333 \cdot 0.5$
$v_{\max} = 1.7 m/s$
$v_{\max} \approx 1.7$ m/sto 2 significant figures

Conclusion

The piston moves fastest through the middle of its stroke, at the angular frequency times half the stroke.

The Ruby

1require_relative "../physics"23question :a_locomotive_piston do4  kind :numerical5  describes "The piston in the cylinder head of a locomotive has a stroke (twice the amplitude) of %{stroke}. If the " \6            "piston moves with simple harmonic motion with an angular frequency of %{angular_frequency}, %{speed}, " \7            "in m/s?"8  given stroke: "1.0 m"9  given angular_frequency: "200 rad/min"10  asks speed: "what is its maximum speed"11end

1require_relative "../questions/a_locomotive_piston"2require_relative "../science/oscillations/simple_harmonic_motion"34solve :a_locomotive_piston, using: SimpleHarmonicMotion do5  objects :piston67  take stroke: piston.stroke8  take angular_frequency: piston.angular_frequency910  speed = piston.maximum_speed11  answer speed:, in: "m/s", figures: 21213  conclusion "The piston moves fastest through the middle of its stroke, at the angular frequency times half " \14             "the stroke."15end

1require_relative "../solutions/a_locomotive_piston"23play_with :a_locomotive_piston do4  title "A Locomotive Piston"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
stroke of ground	$s_{g}$	m	—
amplitude of ground	$A_{g}$	m	0 … ∞
stroke of piston	$s$	m	—
amplitude of piston	$A$	m	0 … ∞
angular frequency of ground	$ω_{g}$	rad/s	0 … ∞
frequency of ground	$ν_{g}$	Hz	0 … ∞
angular frequency of piston	$ω$	rad/s	0 … ∞
frequency of piston	$ν$	Hz	0 … ∞
maximum speed of ground	$v_{max,g}$	m/s	—
maximum speed of piston	$v_{\max}$	m/s	—
maximum acceleration of ground	$a_{max,g}$	m/s²	—
maximum acceleration of piston	$a_{\max}$	m/s²	—
phase constant of piston	$φ$	—	—

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

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(piston)

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(piston)

v_{max,g} = ω_{g} \cdot A_{g}

fastest through the middle(ground)

v_{\max} = ω \cdot A

fastest through the middle(piston)

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(piston)