In: Physics
An astronaut wants to determine the mass of objects she finds in space. On Earth she chooses a spring and attaches a rock to it that weighs 20 N. After placing the rock on a horizontal table, and attaching the spring horizontally to a fixed point, she stretches the rock 2 cm from equilibrium, in the positive direction. She then releases it from rest. After 30 s has passed, she finds that the rock has gone through 10 cycles and is at a new positive maximum value, which is 1 cm from equilibrium. (a) What is the period of the oscillation? (b) What is the spring constant of the spring? (c) What is the time constant of the decay? (d) Find a function that describes the motion, including damping, of the spring and mass system that is only a function of time. Assume time starts when the mass is released. (e) Roughly sketch (1) the position vs time and (2) velocity vs time of this setup. (f) Later, during the return from a mission on Mars (no gravity), she wants to know the mass of an unknown rock she’s found. This rock is attached to the same spring, which is now attached to the wall of the spacecraft. The mass is disturbed from equilibrium and the period of the resulting oscillations is found to be 2 s. What is the ratio of the mass of the new rock to the mass of the rock found on Earth?
(a) time for 10 cycle = 30 s , Hence period of oscillation T = 30/10 = 3 s
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(b) Spring constant k is obtained from ,
...........................(1)
where is angular frequency and m is mass
hence we get ,
.......................(2)
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(c) Amplitude variation is given by, x(t) = xm e-bt/(2m) ................(3)
where b is damping constant
we have xm = initial amplitude = 2 cm and amplitude at time 30 s is 1 cm
hence from eqn.(2) , we get , b = - ( 2 m / t) ln(1/2) = - [ 2 20 / ( 9.8 30 ) ] ln(1/2) = 0.094 kg/s
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(d) x(t) = xm e-bt/(2m) cos( ) .....................(4)
Above equation is position of mass as a function of time. Initial amplitude is xm , b is damping constant, m is mass attached and is angular frequency with damping
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(e) sketch of position Vs Time as given by eqn.(4)
Equation of Velocity as a function of time is obtained from eqn.(3)
....................(5)
sketch of velocity Vs time is given below
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(f) From eqn.(2), we get , m T2
mass ratio = 4/9