Question

The position 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?

Answer 1

Now the stroke is twice the amplitude of the simple harmonic motion.
so the amplitude , A = 1/2 m = 0.5 m
angular frequency of oscillation, ω = 200 rad/min = 200/60 = 10/3 rad/s

we know for a body undergoing simple harmonic motion the velocity is maximum when the body is at mean position or middle position and maximum velocity is given by relation, V = Aω

Where A is the amplitude and ω is the angular frequency of oscillation
So we get maximum velocity of piston as
V = 0.5m × 10/3
= 5/3 m/s
= 1.67 m/s

So we get maximum velocity of the piston is 1.67 m/s.

Answer 2

Answer:

1.66 m/sec.

Explanation:

Here A = 1/2 m

w= 200 rev/ min

Vmax= wA= 200 * 1/2

= 100 m / min.

= 1.66 m / sec.

i hope this helped u...

take the help of photo too, for better understand..