new A = A/10
k and m unchanged
so omega unchanged
T unchanged
E = (1/2) *same k * (new A)^2
= old energy / 100
Completing a lab assignment and was wondering if anyone could look over my work.
A 5.0-g mass is sandwiched between two springs with spring constants k1 and k2. The mass is displaced 10 cm from its equilibrium position and makes sixteen complete oscillations in 1 s with no loss of mechanical energy. Calculate the period of the motion T, the angular frequency w, the sum of the spring constants k1+k2, and the mechanical energy E of the oscillator.
Using the information, I found that the period was equal to 1/16 because the frequency is equal to 1/T.
W=2pi/T
W=100.53 rads/s
W=(k/m)^(1/2)
k=m*w^2
k=50.53 kg/s^2
E=1/2 * k* A^2
E=1/2 * (50.53)*(.10)^2
E=12.76 J
Not really sure how to do the second part of the problem
2. The experiment is repeated, this time displacing the mass only one centimeter from the equilibrium position. Calculate the angular frequency w of the motion and the mechanical energy E of the oscillator.
1 answer
1 answer