Reciprocating engine kinematics — piston in SHM: A steam-engine piston performs approximately simple harmonic motion. The crank rotates at 120 r.p.m. and the stroke length is 2 m (crank radius r = 1 m). Find the linear speed of the piston when it is 0.5 m from the mid-stroke (centre) position.

Difficulty: Medium

Correct Answer: 10.88 m/s

Explanation:


Introduction / Context:
Piston motion in a slider–crank mechanism can be closely modeled as simple harmonic motion (SHM) when the connecting rod is long relative to the crank. This allows fast estimation of piston velocities at given displacements, useful for dynamic force and power calculations in reciprocating machinery.


Given Data / Assumptions:

  • Rotational speed n = 120 r.p.m. → 2 revolutions per second.
  • Angular speed ω = 2πn/60 = 4π rad/s.
  • Stroke = 2 m → crank radius r = 1 m.
  • Displacement from centre x = 0.5 m, SHM approximation is valid.


Concept / Approach:

For SHM with displacement x from mid-stroke, velocity magnitude v is given by v = ω * sqrt(r^2 − x^2). This comes from x = r cosθ and v = r ω sinθ with sinθ = sqrt(1 − (x/r)^2).


Step-by-Step Solution:

Compute ω: ω = 4π ≈ 12.566 rad/s.Use r = 1 m and x = 0.5 m.Evaluate the square root term: sqrt(r^2 − x^2) = sqrt(1 − 0.25) = sqrt(0.75) ≈ 0.8660.Compute velocity: v = ω * sqrt(r^2 − x^2) ≈ 12.566 * 0.8660 ≈ 10.88 m/s.


Verification / Alternative check:

At mid-stroke (x = 0), v would be maximum = r ω = 12.566 m/s. At 0.5 m from centre, the result 10.88 m/s is lower than the maximum and consistent with SHM behavior.


Why Other Options Are Wrong:

5.88 and 8.88 m/s underestimate the SHM velocity at the given position; 12.88 and 14.88 m/s exceed the maximum possible r ω for the data.


Common Pitfalls:

Using uniform linear motion formulas; forgetting to convert r.p.m. to rad/s; mixing stroke and radius.


Final Answer:

10.88 m/s

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