- #1
Samuelriesterer
- 110
- 0
1. The problem statement, equations, and work done:
[A] Two identical spring and mass oscillators are set in motion in perpendicular directions. The masses are each 4.0 kg and the spring constants are 196 N/m.
[1] Calculate the angular frequency and the period of the oscillators.
##\omega = \sqrt{\frac{k}{m}} = .143 rad/sec##
##T = \frac{2\pi}{\omega} = 43.938 s##
[2] The oscillators are damped such that the amplitude decreases by 20% in 140 seconds. Determine the coefficient in the damping force (b in F = -bv) and the time constant (τ) for the energy decay ( E= Eo e-t/τ)
##A_{\% decrease\hspace{1 mm}per\hspace{1 mm}cycle} = \frac{A_{total\hspace{1 mm}decrease} \hspace{1 mm}X\hspace{1 mm}time}{T} = \frac{.20 X 140 s}{43.938 s} = 63.726 \%##
##A = A_0 - A_{\%decrease\hspace{1 mm}per\hspace{1 mm}cycle} = A_0 e^{\frac{-bT}{2m}} →##
##1 - .63726 = 1\hspace{1 mm}X\hspace{1 mm}e^{\frac{-b(43.938 s)}{2(4 kg)}} →##
##b = .184636 kg/s##
##\tau = \frac{m}{b} = \frac{.18636 kg/s}{4 kg} = 21.66 s##
[3] A second observer passes at 0.6c, traveling in the direction one of the oscillators moves. Write the force law for the spring in the frame where one end is stationary (as in parts 1 and 2). Then transform this to the second observer's frame by finding the inertia and stretch in the second frame. Do this twice, once for each oscillator.
##F_{stationary\hspace{1 mm}reference} = ma = -k\Delta x##
<This is where I am confused. How to I convert this to second observer's frame by finding the inertia and stretch?>
[4] Using your force law in [3], derive the expressions for the angular frequency and the period of the oscillators in the second frame.
[5] Now compare the periods you calculated to the result of a time dilation calculation
[A] Two identical spring and mass oscillators are set in motion in perpendicular directions. The masses are each 4.0 kg and the spring constants are 196 N/m.
[1] Calculate the angular frequency and the period of the oscillators.
##\omega = \sqrt{\frac{k}{m}} = .143 rad/sec##
##T = \frac{2\pi}{\omega} = 43.938 s##
[2] The oscillators are damped such that the amplitude decreases by 20% in 140 seconds. Determine the coefficient in the damping force (b in F = -bv) and the time constant (τ) for the energy decay ( E= Eo e-t/τ)
##A_{\% decrease\hspace{1 mm}per\hspace{1 mm}cycle} = \frac{A_{total\hspace{1 mm}decrease} \hspace{1 mm}X\hspace{1 mm}time}{T} = \frac{.20 X 140 s}{43.938 s} = 63.726 \%##
##A = A_0 - A_{\%decrease\hspace{1 mm}per\hspace{1 mm}cycle} = A_0 e^{\frac{-bT}{2m}} →##
##1 - .63726 = 1\hspace{1 mm}X\hspace{1 mm}e^{\frac{-b(43.938 s)}{2(4 kg)}} →##
##b = .184636 kg/s##
##\tau = \frac{m}{b} = \frac{.18636 kg/s}{4 kg} = 21.66 s##
[3] A second observer passes at 0.6c, traveling in the direction one of the oscillators moves. Write the force law for the spring in the frame where one end is stationary (as in parts 1 and 2). Then transform this to the second observer's frame by finding the inertia and stretch in the second frame. Do this twice, once for each oscillator.
##F_{stationary\hspace{1 mm}reference} = ma = -k\Delta x##
<This is where I am confused. How to I convert this to second observer's frame by finding the inertia and stretch?>
[4] Using your force law in [3], derive the expressions for the angular frequency and the period of the oscillators in the second frame.
[5] Now compare the periods you calculated to the result of a time dilation calculation