A mass oscillating up and down on the bottom of a spring can be modeled as harmonic motion. if they weight is displaced a maximu
m of 7 cm and it takes 4.0 sec to complete one cycle to the nearest hundredth of a cm how far above or below the resting position is the weight after 1.15 seconds ?
1 answer:
9514 1404 393
Answer:
1.63 cm (across the centerline from release)
Step-by-step explanation:
If we assume time starts counting when we release the weight from its fully-extended downward position, then the position at 1.15 seconds can be found from ...
h(t) = -7cos(2πt/4)
h(1.15) = -7cos(π·1.15/2) = -7(-0.233445) ≈ 1.63412 . . . cm
That is, 1.15 seconds after the weight is released from below the resting position, it will be 1.63 cm above the resting position.
__
If it is released from <em>above</em> the resting position, it will be 1.63 cm <em>below</em> the resting position at t=1.15 seconds.
You might be interested in
We know that : (a - b)(a + b) = a² - b²
We know that : 1 - sin²x = cos²x
We know that : sec²x = 1 + tan²x
Answer:
Step-by-step explanation:
1. Distribute the X across the equation:
2.
3.
3.