Analyzing Newton\'s 2nd Law for a mass spring system, we found alpha x = -k/m x.
ID: 1467639 • Letter: A
Question
Analyzing Newton's 2nd Law for a mass spring system, we found alpha x = -k/m x. Comparing this to the x-component of uniform circular motion, we found as a possible solution for the above equation x = A cos (omega t) V x = -omega A sin(omega t) alpha x = -omega^2 A cos(omega t) with omega = square root k/m, and A the amplitude (maximum displacement from equilibrium). Consider an oscillator with mass 0.383 kg and spring constant 13 6 N/m If the amplitude of the oscillation is 6 61 cm. what is the maximum speed? (Note, you might already have solved similar problems employing conservation of energy Here, Td like you to find the answer using the equations of motion.) Answer in Consider an oscillator with frequency 27 Hz If at t=0 the oscillator is at a maximum displacement from the equilibrium Consider an oscillator with mass 0.32 kg and spring constant 12 8 N/kg The oscillator is displaced from the equilibrium position by +5.47 cm and released After what same, is the oscillator for the first time at a displacement of -2 17 cm?Explanation / Answer
w = (k/m)^1/2
= (13.6/0.383)^1/2
=
Vmax = w*a*sin90
= 0.0661* 5.95895
= 0.394 m/s
Cannot see the question 2 completely
3) w = (12.8/0.32)^1/2
x = A cos(wt)
At t= 0
x = 5.47
Hence
5.47 = A *cos0
A = 5.47
Now displacement is -2.17 Hence
-2.17 = 5.47 * cos ( 6.32456*t)
t = 0.3129 sec