1) When a pendulum 0.5m long swings back and forth, its angular displacement Ɵ from rest position, in radians is given by Ɵ=1/4sin((pi/2)t), where t is the time, in seconds. At what time(s) during the first 4 s is the pendulum displaced 1 cm vertically above its rest position? (assume the pendulum is at its rest position at t=0).
my first equation should have been
cosØ = 49/50 = .98
Why did the pendulum bring a tape measure to the party? Because it wanted to measure its vertical displacement, of course! Let's swing into action and solve this problem.
To find when the pendulum is displaced 1 cm above its rest position, we need to set the angular displacement equation equal to 1 cm. But before we do that, we need to convert the length of the pendulum from meters to centimeters because apparently, the pendulum enjoys metric conversions as much as I enjoy cracking jokes!
Given that the pendulum is 0.5m long, it can also be expressed as 50 cm. So now the equation becomes:
Ɵ = 1/4sin((pi/2)t) = 1 cm
Alrighty then, to find when this happens within the first 4 seconds (because the pendulum has a fantastic sense of timing), we can set up an equation:
1/4sin((pi/2)t) = 1
Now all we have to do is solve for t, the time in seconds. So here's the plan: we'll multiply both sides by 4 to get sin((pi/2)t) = 4, then take the inverse sine on both sides to eliminate the sin, and finally divide by (pi/2) to get t by itself.
But don't worry, I promise to keep the math jokes to a minimum during this calculation. Math can be daunting, but we'll make it fun!
sin((pi/2)t) = 4
Taking the inverse sine on both sides gives us:
(pi/2)t = sin^(-1)(4)
Dividing by (pi/2) yields:
t = sin^(-1)(4)/(pi/2)
"Wait a minute," you might say. "There's something fishy going on here! The arcsine of 4 isn't even a real number!"
Well, you're absolutely right! Turns out, the equation has no real solution in this case. So no time within the first 4 seconds will the pendulum be displaced 1 cm above its rest position.
Looks like this pendulum prefers to stay closer to the party in the rest position. Sorry for spinning the math gears and not finding a solution, but hey, at least we had some fun along the way, right?
To find the time(s) during the first 4 seconds when the pendulum is displaced 1 cm above its rest position, we need to solve the given equation for Ɵ.
Given: Ɵ = (1/4) sin((π/2)t)
We know that the pendulum is displaced 1 cm above the rest position when Ɵ = 0.01 meters (since 1 cm is equal to 0.01 meters).
So, we need to solve the equation Ɵ = (1/4) sin((π/2)t) = 0.01 meters for t.
Here's how you can solve it:
1. Start by multiplying both sides of the equation by 4 to get rid of the fraction:
4Ɵ = sin((π/2)t)
2. Next, take the inverse sine (or arcsine) of both sides of the equation to isolate t:
arcsin(4Ɵ) = (π/2)t
3. Divide both sides of the equation by (π/2) to solve for t:
t = (2/π) * arcsin(4Ɵ)
Now, substitute Ɵ = 0.01 meters into the equation to find the value(s) of t when the pendulum is displaced 1 cm above its rest position during the first 4 seconds.
Finally, evaluate the equation t = (2/π) * arcsin(4Ɵ) for Ɵ = 0.01 meters to find the time(s) during the first 4 seconds when the pendulum is displaced 1 cm vertically above its rest position.
The second answer is wrong it should be 3.41seconds instead of 2.55s
Make a situation diagram where you have a triangle with the vertical angle as Ø, the hypotenuse 50 cm (the pendulum) and a vertical leg of 49 cm
so we would get
cosØ - 49/50 = .98
then Ø = .20033.. (I stored that in my calculator to keep accuracy)
so we have
(1/4)sin(πt/2) = .20033..
sin (πt/2) = .801339...
πt/2 = .9295...
t = .5917577 or π - .59175.. = 2.5498...
the pendulum will be 1 cm above its rest position at
t = appr .59 seconds or at t = appr 2.55 seconds