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.

(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?

Tried the formula 2pi(m/k)^2 but not working, don't know what to do... help

Damping is exponential.

F(h)= a * e^-bt/2m * sin(wt+ phase)

with b the damping constant, which comes from a differential equation. I think you have the units wrong. The units should be mass/time

Hmmm. I see you have N sec/m which is mass*m/sec^2*sec/mass...= sec/kg Ok, you can use those units.

solution:

1/2= e^-.7/5.8 t
take the ln of each side.
-.693=.7/5.8 t solve for t

You need the formula that relates the damping constant "b" (retarding force divided by speed) to the exponential damping RATE (lambda) for damped harmonic oscillations. This must have been in your reading material somewhere.

If not, see "underdamped oscillations" in
http://physics.ucsc.edu/~josh/6A/book/harmonic/node18.html

The exponential damping time constant is
lambda = b/2m = 0.12 sec-1

e^-lambda*t = 1/2 when
-lambda*t = -0.693
t = 5.8 seconds

To solve this problem, we need to use the equation of motion for a damped harmonic oscillator. The equation is:

m * x'' + b * x' + k * x = 0

where m is the mass, x is the displacement from the equilibrium position, b is the damping constant, and k is the spring constant.

In this case, the student stops moving her hand, so there is no longer an external force acting on the system. This means that the damping term (b * x') becomes zero. The equation simplifies to:

m * x'' + k * x = 0

Given the values in the problem, we have m = 2.9 kg and k = 93 N/m. We need to solve for the time it takes for the amplitude of the mass to reach one-half its maximum amplitude.

To solve this, we can use the equation for the period of a damped harmonic oscillator:

T = 2π / (ω * sqrt(1 - (b / (2 * m * ω))^2))

where T is the period, ω is the angular frequency, and b is the damping constant.

The angular frequency can be calculated using the formula:

ω = sqrt(k / m)

Substituting the values into the formulas, we have:

ω = sqrt(93 N/m / 2.9 kg) ≈ 5.000 rad/s

Plugging this value into the period equation, we get:

T = 2π / (5.000 rad/s * sqrt(1 - (0.7 N s/m / (2 * 2.9 kg * 5.000 rad/s))^2))

Simplifying further, we have:

T = 2π / (5.000 rad/s * sqrt(1 - (0.7 N s/m / 29.00 N s/m)^2))

T = 2π / (5.000 rad/s * sqrt(1 - 0.01622))

T = 2π / (5.000 rad/s * 0.998)

T ≈ 2π / 4.994 rad/s

T ≈ 1.26 s

So it will take approximately 1.26 seconds for the amplitude of the mass to reach one-half its maximum amplitude after the student stops moving her hand.