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3. An astronaut lands on an alien planet. He places a pendulum (L = 0.200 m) on the surface and sets it in simple harmonic motion, as shown in this graph.

Answer the following questions:
a. What is the period and frequency of the pendulum’s motion?
b. How many seconds out of phase with the displacements shown would graphs of the velocity and acceleration be?
c. What is the acceleration due to gravity on the surface of the planet in m/s2? Determine the number of g-forces.
Show any necessary calculations.
Answer:

3 An Astronaut Lands On An Alien Planet He Places A Pendulum L 0200 M On The Surface And Sets It In Simple Harmonic Motion As Shown In This Graph Answer The Fol class=

Sagot :

AL2006

All this work for 5 points is a big rip-off, but I'm going to try it anyway,
for myself, just to see if I can do it.

a).  From the graph, I count 9 complete cycles in 5 seconds.
So the frequency is  9/5  =  1.8 per second = 1.8 Hz.
The period is  1/frequency = 5sec / 9 cycles = 0.555... sec .

b).  Velocity = first derivative of displacement = 1/4 cycle behind =
       5/36 sec delayed with respect to displacement = 0.13888... sec.

       Acceleration = first derivative of velocity = 2nd derivative of displacement =
       1/4 cycle behind velocity = 1/2 cycle behind displacement =
       5/36 (0.13888...) sec delayed with respect to velocity, =
       5/18 (0.2777...) sec delayed with respect to displacement.

c).  For small swing angles, the period of an ideal pendulum anywhere is 

                 T = 2pi √(length / local gravity)  .

The astronaut has already done the pendulum and transmitted the data to us, so
we can use his data and this formula to calculate the local gravity where he is.

           P = 2pi √(length / local gravity)

            5/9 sec = 2π √(0.2m / gravity)

         √(0.2m / gravity) =  5/9sec / 2π

Take the reciprocal of each side:    √(gravity) / √(0.2) =  18π / 5

Multiply each side by  √(0.2):    √(gravity) = 18π √(0.2) / 5

Square both sides:      Gravity = (324 π²) (0.2) / 25  =  25.582 m/sec²

This is about  2.608 times  the Earth's gravity.  So it should not surprise us
that the astronaut got fed up playing with his pendulum after only 5 seconds,
and went back to his landing capsule to lie down.

Note:
Even though it's a highly classified secret, closely guarded for reasons of
national security and all that stuff, we can be pretty sure that our man has
landed on Jupiter.  His data and our calculations have produced a value of
25.582 m/s²  for the acceleration of gravity where he is.  This compares with
the value of  24.79 m/s²  measured by previous robotic space-probe missions
to Jupiter.  The difference is less than 3.1% .

The only remaining questions are:

--  How is he managing to sit on top of the planet's gaseous envelope, playing
with his pendulum and staying in radio contact with us, without falling in ?

--  According to the graph, the pendulum was practically at zero displacement
when he released it at Time=0, and the first thing it did when he let go of it
was to swing out to 0.04 radians before turning back.  This troubles me.
Either the pendulum was already swinging before Time=0 on this graph,
or else his data have been seriously doctored.


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