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Question : 14 of 25
Marks: +1, -0
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−1200\,\text{rad}\,\text{min}^{-1}, what is its maximum speed?
Solution:  
Stroke of piston = 2 times the amplitude. Let A = amplitude, Stroke = 1 m (given)
∴1=2A\therefore 1=2A or A=12 mA=\frac{1}{2}\,\text{m}
Now,  vmax⁡=ωA=200×12\text{Now},\; v_{\max}=\omega A=200\times\frac{1}{2}
=100 m/min=10060 m s−1=100\,\text{m/min} = \frac{100}{60}\,\text{m}\,\text{s}^{-1}
=53 m s−1=1.67 m s−1=\frac{5}{3}\,\text{m}\,\text{s}^{-1}=1.67\,\text{m}\,\text{s}^{-1}
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