A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass
uming the motion is vertical calculate the amplitude of the ship's motion, if the scale reading of the machine varies between limits of 55.0kg and 65.0kg.
Well, first of all, one who is sufficiently educated to deal with solving this exercise is also sufficiently well informed to know that a weighing machine, or "scale", should not be calibrated in units of "kg" ... a unit of mass, not force. We know that the man's mass doesn't change, and the spectre of a readout in kg that is oscillating is totally bogus.
If the mass of the man standing on the weighing machine is 60kg, then on level, dry land on Earth, or on the deck of a ship in calm seas on Earth, the weighing machine will display his weight as 588 newtons or as 132.3 pounds. That's also the reading as the deck of the ship executes simple harmonic motion, at the points where the vertical acceleration is zero.
If the deck of the ship is bobbing vertically in simple harmonic motion with amplitude of M and period of 15 sec, then its vertical position is
y(t) = y₀ + M sin(2π t/15) .
The vertical speed of the deck is y'(t) = M (2π/15) cos(2π t/15)
and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)
= - (4 π² M / 15²) sin(2π t/15)
= - 0.1755 M sin(2π t/15) .
There's the important number ... the 0.1755 M. That's the peak acceleration. From here, the problem is a piece-o-cake.
The net vertical force on the intrepid sailor ... the guy standing on the bathroom scale out on the deck of the ship that's "bobbing" on the high seas ... is (the force of gravity) + (the force causing him to 'bob' harmonically with peak acceleration of 0.1755 x amplitude).
At the instant of peak acceleration, the weighing machine thinks that the load upon it is a mass of 65kg, when in reality it's only 60kg. The weight of 60kg = 588 newtons. The weight of 65kg = 637 newtons. The scale has to push on him with an extra (637 - 588) = 49 newtons in order to accelerate him faster than gravity.
Now I'm going to wave my hands in the air a bit:
Apparent weight = (apparent mass) x (real acceleration of gravity)
(Apparent mass) = (65/60) = 1.08333 x real mass.
Apparent 'gravity' = 1.08333 x real acceleration of gravity.
The increase ... the 0.08333 ... is the 'extra' acceleration that's due to the bobbing of the deck.
0.08333 G = 0.1755 M
The 'M' is what we need to find.
Divide each side by 0.1755 : M = (0.08333 / 0.1755) G
That result fills me with an overwhelming sense of no-confidence. But I'm in my office, supposedly working, so I must leave it to others to analyze my work and point out its many flaws. In any case, my conscience is clear ... I do feel that I've put in a good 5-points-worth of work on this problem, even if the answer is wrong .
When rounding you look at the number right after the one you want to round. The tenth place is the one right after the decimal. If the number after it is more than five it goes up by one, it it's less than five it stays the same. So since the number after the 8 is a 3 it stays the same.
