- #1
gnome
- 1,041
- 1
"simple" harmonic motion
Trying to solve a harmonic motion problem (mass oscillating on a spring), given:
amplitude=.100 m
k=8.00 N/m
mass=.5 kg
to solve for time elapsed for the mass to travel from x=0.00 to x=.080 m.
It's easy to solve for w (omega) = sqrt(8/.5) = 4, and using the equation for shm from my physics textbook I got this equation:
.08 = .1 cos(4t-pi/2)
and then
t = ((arccos 0.8) + pi/2)/4 = .554 sec. which unfortunately is wrong. This turns out to be the time it takes for the mass to go past .08 m, past its maximum displacement of .1 m, and BACK to .08 m.
Now, this equation is equivalent:
.08 = .1 sin(4t)
Check it...the graphs are exactly the same. But this one solves to
t = .232 sec. which is the answer given by the book, and tracing the graph confirms that this is the correct answer.
So, there must be a way to solve the first equation [.08 = .1 cos(4t-pi/2)] to get the correct answer of .232 sec., but I can't see it.
What am I missing?
Trying to solve a harmonic motion problem (mass oscillating on a spring), given:
amplitude=.100 m
k=8.00 N/m
mass=.5 kg
to solve for time elapsed for the mass to travel from x=0.00 to x=.080 m.
It's easy to solve for w (omega) = sqrt(8/.5) = 4, and using the equation for shm from my physics textbook I got this equation:
.08 = .1 cos(4t-pi/2)
and then
t = ((arccos 0.8) + pi/2)/4 = .554 sec. which unfortunately is wrong. This turns out to be the time it takes for the mass to go past .08 m, past its maximum displacement of .1 m, and BACK to .08 m.
Now, this equation is equivalent:
.08 = .1 sin(4t)
Check it...the graphs are exactly the same. But this one solves to
t = .232 sec. which is the answer given by the book, and tracing the graph confirms that this is the correct answer.
So, there must be a way to solve the first equation [.08 = .1 cos(4t-pi/2)] to get the correct answer of .232 sec., but I can't see it.
What am I missing?