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Exercises · 13.14

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?

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The piston’s maximum speed in SHM is the product of amplitude and angular frequency. With stroke 1.0 m (so amplitude 0.5 m) and ω=200\omega = 200 rad/min, the maximum speed is vmax=100v_{\text{max}} = 100 m/min.

Why this works — the SHM speed story

In simple harmonic motion, the position of the piston varies sinusoidally: x(t)=Acos⁡(ωt+ϕ)x(t) = A \cos(\omega t + \phi). The speed is the time derivative of position. Because the cosine function changes fastest when it crosses zero (the equilibrium point), that’s where the piston’s speed peaks. The maximum speed is simply the amplitude times the angular frequency — a direct consequence of differentiating a sinusoid.

The stroke is the total distance the piston travels from one extreme to the other — that’s twice the amplitude. So amplitude is half the stroke. Angular frequency ω\omega is given in rad/min, so the speed will come out in metres per minute.

Step-by-step

  1. Find the amplitude from the stroke. Stroke = 2A=1.02A = 1.0 m, so

A=1.02=0.5 m.A = \frac{1.0}{2} = 0.5 \text{ m}.

  1. Recall the formula for maximum speed in SHM. For x=Acos⁡(ωt+ϕ)x = A \cos(\omega t + \phi),

v=dxdt=−Aωsin⁡(ωt+ϕ).v = \frac{dx}{dt} = -A\omega \sin(\omega t + \phi).

The sine function varies between −1-1 and 11, so the magnitude of speed is at most AωA\omega. Hence

vmax=Aω.v_{\text{max}} = A\omega.

vmax=Aωv_{\text{max}} = A\omega

  1. Plug in the numbers. A=0.5A = 0.5 m, ω=200\omega = 200 rad/min. …

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