An artificial satellite orbits Earth at a speed of 7800 m/s and a height of 200 km above Earth's surface. The satellite experien
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(a) 4.06 cm
In a simple harmonic motion, the displacement is written as
(1)
where
A is the amplitude
is the angular frequency
is the phase
t is the time
The displacement of the piston in the problem is given by
(2)
By putting t=0 in the formula, we find the position of the piston at t=0:
(b) -14.69 cm/s
In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:
(3)
Differentiating eq.(2), we find
And substituting t=0, we find the velocity at time t=0:
(c) -101.13 cm/s^2
In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:
Differentiating eq.(3), we find
And substituting t=0, we find the acceleration at time t=0:
(d) 5.00 cm, 1.26 s
By comparing eq.(1) and (2), we notice immediately that the amplitude is
A = 5.00 cm
For the period, we have to start from the relationship between angular frequency and period T:
Using and solving for T, we find