a 0.300 kg mass is attached to a 26.6 n/m spring. it is pulled 0.120 m and released. what is the speed of the mass when it is .0600 m form equilibrium?( Unit=m/s)
- RealProLv 72 years agoFavorite Answer
The sum of kinetic and potential energy is conserved in a simple harmonic oscillator.
We are told the mass is starting at an maximum elongation A=0.12 m with 0 speed.
So total energy at that point is entirely held in the potential energy kA^2 / 2, because the kinetic energy is 0.
E = kA^2 / 2 + 0 = 0.1915 J
When the mass is getting pulled towards equillibrium point, this stored potential energy is being turned into kinetic energy of the mass.
At x=0.06 m potential energy is only
Ep = kx^2 / 2 = 0.04788 J
So the missing part (0.1915 - 0.4788 = 0.14364) has gone to kinetic.
mv^2 / 2 = 0.14364 J
v = 0.979 m/s to 3 sig digits.
In short you can always apply the rule x^2 + (m/k)v^2 = A^2
Where A is the amplitude of the motion the body is going through.
- Anonymous2 years ago