Solving Trigonometric Equations

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The discussion focuses on solving two trigonometric equations related to a pendulum's motion and the current in a household appliance. For the pendulum, participants suggest using trigonometric functions to determine when it is displaced 1 cm above its rest position, emphasizing the need to consider the maximum height and the correct angles. In the second problem, there is confusion about calculating the rate of change of current at t=1s, with one user initially arriving at zero and others correcting this by suggesting a more accurate approach using limits. The importance of ensuring calculators are in radian mode and tracking negatives is highlighted. Overall, the conversation revolves around clarifying the methods for solving these trigonometric problems.
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Homework Statement



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 0).

2) the current in a household appliance varies according to the equation A=5sin120pit, where A is the current in amperes, and t is the time, in seconds. at what rate is hte current changing at t=1s?





The Attempt at a Solution


1) I'm not sure how to approach/solve this question

2) i got 0, can anyone confirm?
 
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BlackOut07 said:

Homework Statement



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 0).

2) the current in a household appliance varies according to the equation A=5sin120pit, where A is the current in amperes, and t is the time, in seconds. at what rate is hte current changing at t=1s?





The Attempt at a Solution


1) I'm not sure how to approach/solve this question
Start by sketching a graph. At what points on the graph is the height above the starting position 1 cm?
BlackOut07 said:
2) i got 0, can anyone confirm?
Show us how you got 0.
 
a=(5sin120pi(1))-(5sin120pi(0.999))/(1-0.999)
=0?? doesn't make sense though
 
BlackOut07 said:
a=(5sin120pi(1))-(5sin120pi(0.999))/(1-0.999)
=0?? doesn't make sense though

What you have calculated here is the current at t = 1. Is the question not asking for the rate of change at t = 1?
 
gordonj005 said:
What you have calculated here is the current at t = 1. Is the question not asking for the rate of change at t = 1?

so would it be
a=(5sin120pi(1.001))-(5sin120pi(0.999))/(1.001-0.999)
 
Yes, that's a pretty close approximation and will give you an answer within 2.35 % of the exact answer. Question, have you ever done any calculus before? (and yes, I do realize this is the precalculus section)
 
gordonj005 said:
Yes, that's a pretty close approximation. Question, have you ever done any calculus before?

no, i have it second semester.

and the answer is STILL 0 :/
 
I assure you the answer is not zero, make sure your calculator is in radian mode, and make sure you keep track of your negatives.
 
BlackOut07 said:

Homework Statement



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 0).

the maximum of this graph is 0.25 :/
gordonj005 said:
I assure you the answer is not zero, make sure your calculator is in radian mode, and make sure you keep track of your negatives.

i got the answer as a=-328.365.
 
  • #10
For the pendulum, think in terms of trigonometric functions. If you set the rest position to (0, 0) and you know the radius is 0.5 m, for what values of \theta will the height be 1 cm? Once you figure that out, you can find the times fairly easily.

Ok so:

m = \frac{5 \sin{120\pi 1.001} - 5 \sin{120\pi 0.999}}{0.002}
m = 2500(\sin{120.12 \pi} - \sin{119.88 \pi})

where \sin{120.12 \pi} \approx 0.368 and \sin{120\pi 0.999} \approx -0.368. I think if you try again you'll get the right answer.
 
  • #11
i still don't understand to be honest :/
 
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