what happens to the maximum velocity if you double both Amplitude and period of a SHM?
the equation is
x(t) = A cos (wt + f)
v(t) = -A w sin (wt + f)
f is phase constant, w = angular velocity
i know if i double A and T, the first part of the v(t) equation will be like:
-2A *...
waiiiit a second..period 2.4 s that's going forward and back...
cut that by half, that's going one way trip
cut that by half again...would it be..could it be..AT THE EQILIBRIUM POINT!?
2.4/4 = 0.6
so the time to arrive at the equilibrium point is...0.6!
now i see, the bottom equation...
i see what you're saying
the maximum speed occurs when it is at its equlibrium point, x = 0 right?
using the first derivitive i get velocity equation, which is:
v(t) = - A w sin (wt) w = 2.61
-46 = - A*2.61 sin(2.61*t) <---this is at the eqi. point, going to the left
but i still have 2...
An air-track glider is attached to a spring. The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period of 2.40 s and a maximum speed of 46.0 cm/s.
What is the amplitude of the oscillation?
i tryed solving for t by using the equation x(t) = A cos...
k ktd
here's how you do it lol
see 2 blocks as one 4500 / 3000 = 1.5 (both's acc)
f = ma, so force of car on truck = 1.5 * 1000 (car's mass) = 1500
now, the car's wheel produces 4500 force, we said 1500's car pushing truck, so where did the other 3000 go??
its truck pushing back
<-----...
A 1000 kg car pushes a 2000 kg truck that has a dead battery. When the driver steps on the accelerator, the drive wheels of the car push against the ground with a force of 4500 N.
What is the magnitude of the force of the car on the truck?
What is the magnitude of the force of the truck on the...