Say x = A sin wt
then
v = A w cos w t
max speed is when cos w t = 1 or -1
That is when sin w t = 0
which is when x = 0
1/2 max is when cos w t = 1/2
We need to know sin wt when cos w t = 1/2
BUT we know that cos^2+sin^2 = 1 (trig)
so
(1/2)(1/2) + sin^2 wt = 1
sin^2 wt = 3/4
sin wt = .866
so
x = .866 A at half max speed
-------------------------
Now PE = (1/2) k x^2
max PE = .5 k A^2
when is PE = (1/2)(1/2) k A^2 ???
(1/2) k x^2 = (1/2)(1/2) k A^2
x^2 = .5 A^2
x = .707 A
A mass m is attached to a spring of constant k. The spring oscillates in simple harmonic motion (amplitude A). In terms of A:
a) At what displacement from equilibrium is the speed half of the max value and b) at what displacement from equilibrium is the potential energy half of the max value? Why is letter b greater than letter a?
Any help is appreciated.
3 answers
Ah, I really appreciate that. Thank you.
The kinetic energy + potential energy is constant. Each goes to zero when the other is max.
Therefore when the PE is 1/2 max, the KE will also be (1/2) max
the half PE occurs when x = .707 A, part b
so the half KE has to happen at x = .707 A as well
that means (1/2) m v^2 = (1/2)(1/2)m Vmax^2
at x = .707 A
v^2 = .5 Vmax^2 at x = .707 A
so
v = .707 Vmax at .707 A
THEREFORE v does not reach .5 Vmax until further out than .707 A (speed max at x = 0, and speed 0 at x = A)
SO - I claim that the answer to b is SMALLER than the answer to a
Therefore when the PE is 1/2 max, the KE will also be (1/2) max
the half PE occurs when x = .707 A, part b
so the half KE has to happen at x = .707 A as well
that means (1/2) m v^2 = (1/2)(1/2)m Vmax^2
at x = .707 A
v^2 = .5 Vmax^2 at x = .707 A
so
v = .707 Vmax at .707 A
THEREFORE v does not reach .5 Vmax until further out than .707 A (speed max at x = 0, and speed 0 at x = A)
SO - I claim that the answer to b is SMALLER than the answer to a