A student gets a 2.9 kg mass to oscillate up and down on bottom of a light vertical spring by pulling her hand up and down on the top end of the spring. The spring is a real spring with a spring constant of 93 N/m and a damping constant of 0.7 N sec/m.
(a) At what approximate frequency should the student move her hand up and down to get the maximum motion from the mass with the minimum motion of her hand?
fhando = /sec *
.90 OK
(b) The student now stops moving her hand and the mass slowly comes to rest. How long after she stop shaking her hand will it take for the amplitude of the mass to reach one half its maximum amplitude?
t1/2= seconds
i got .90 for part a , cant figure out b
idk lol
To solve part (a) of the problem, we need to find the approximate frequency at which the student should move her hand to achieve maximum motion of the mass and minimum motion of her hand.
The frequency of oscillation for a mass-spring system can be calculated using the formula:
f = (1 / 2π) * sqrt(k / m)
where f is the frequency, k is the spring constant, and m is the mass. In this case, k = 93 N/m and m = 2.9 kg.
Plugging in these values, we can calculate the frequency:
f = (1 / 2π) * sqrt(93 / 2.9) ≈ 2.58 Hz
Therefore, the student should move her hand up and down at approximately 2.58 Hz to achieve the desired motion.
For part (b), we need to find the time it takes for the amplitude of the mass to decrease to half its maximum value after the student stops moving her hand. This can be determined using the damped oscillation equation:
A(t) = A0 * exp(-at / (2m))
where A(t) is the amplitude at time t, A0 is the initial amplitude, a is the damping constant, and m is the mass. In this case, A0 is the maximum amplitude of the mass and is not given. Let's represent it as Amax.
We're looking for the time t1/2 when the amplitude becomes half its maximum value:
A(t1/2) = (1/2) * Amax
Substituting this into the equation, we have:
(1/2) * Amax = Amax * exp(-a * t1/2 / (2m))
Now, we can solve for t1/2:
(1/2) = exp(-a * t1/2 / (2m))
Taking the natural logarithm (ln) of both sides:
ln(1/2) = -a * t1/2 / (2m)
Rearranging:
t1/2 = -(2m / a) * ln(1/2)
Plugging in the values for m = 2.9 kg and a = 0.7 N sec/m, we can calculate t1/2:
t1/2 = -(2 * 2.9 / 0.7) * ln(1/2) ≈ 6.22 seconds
Therefore, it would take approximately 6.22 seconds for the amplitude of the mass to reach one-half its maximum amplitude after the student stops shaking her hand.