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प्रश्न
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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उत्तर १
Stroke of piston = 2 times the amplitude
Let A = amplitude,stroke = 1 m
`:. => A = 1/2 m`
Angular frequency, ` omega = 200 "rad/min"`
`V_"max" = ?`
We know that the maximum speed of the block when the amplitude is A
`V_"max" = omegaA = 200 xx 1/2 = 100 "m.min"`
`= 100/60 = 5/3 ms6(-1)= 1.67 ms^(-1)`
उत्तर २
Angular frequency of the piston, ω = 200 rad/ min.
Stroke = 1.0 m
Amplitude, `A = 1.0/ 2= 0.5 m`
The maximum speed (vmax) of the piston is give by the relation:
`V_"max" = Aomega`
`= 200 xx 0.5 = 100 "m/min"`
संबंधित प्रश्न
The periodic time of a linear harmonic oscillator is 2π second, with maximum displacement of 1 cm. If the particle starts from extreme position, find the displacement of the particle after π/3 seconds.
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