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Alecsey [184]
3 years ago
10

Starting at its rightmost position, it takes 2 seconds for the pendulum of a grandfather clock to swing a horizontal distance of

18 inches from right to left and 2 seconds for the pendulum to swing back from left to right. Which of the following equations models d, the horizontal distance in inches of the pendulum from the center as a function of time, t, in seconds? Assume that right of center is a positive distance and left of center is a negative distance.
a) d=9cos(pi/4 t)
b) d=9cos(pi/2 t)
c) d=18cos(pi/4 t)
d) d=18cos(pi/2 t)
Physics
2 answers:
sdas [7]3 years ago
8 0

Answer:

The correct answer is b,  x = 9 cos (pi / 2 t)

Explanation:

The equation that describes a simple pendulum is

             θ  = θ₀  cos (wt + φ)

The angle is measured is radians

            θ = x / L

We replace

           d / L = x₀ / L cos (wt + φ)

            x₀ = 9 in

         

We replace

             d = 9 cos (wt + φ)

Angular velocity is related to frequency and period.

           w = 2π f = 2π / T

The period is the time of a complete oscillation T = 4 s

           w =2π / 4

           w = π / 2

Let's replace

             x = 9 cos (π/2 t + φ)

As the system is released from the root x = x₀ for t = 0 s

              x₀ = x₀ cos φ

             Cos φ = 1

             φ = 0°

The final equation is

             x = 9 cos (pi / 2 t)

The correct answer is b

N76 [4]3 years ago
6 0

Answer:

b) d=9cos(pi/2 t)

Explanation:

This is a cosine function in the such as: y = a cos bt...

a = (maximum distance - minimum distance)/2:

a = (max - min)/2

maximum distance = 18 inches minimum distance = 0

a = (18 - 0)/2 = 18/2 = 9

That is a = 9

To solve for b, similar to the period:

The period in radians:

P = 2pi/b, which is the amount of time it takes to revolve one full cycle...

multiply that time by time from minimum to maximum, 2 to give the period as 2 × 2 = 4.

Thus to find b..

4 = 2pi/b

4b = 2pi/b × b

4b = 2pi

or

b = pi/2

Also

a = 9

b = pi/2

So our cosine function is:

d = 9cos((pi/2)t)

Hence the equation that models d

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This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity

        sin θ = v_{oy} / vo

        cos θ = v₀ₓ / vo

         v_{oy} = v_{o} sin θ

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          x = v₀ₓ t

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          0 = v₀ sin θ - gt

          t = v_{o} sin θ / g

         

we substitute

       x = v₀ cos θ (2 v_{o} sin θ / g)

       x = v₀² /g      2 cos θ sin θ

       x = v₀² sin 2θ / g

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b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time

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          v_{y} = v_{oy} - gt

          v_{y} = 0

           t = v_{oy} / g

           t = v₀ sin θ / g

as the time to get on and off is the same the total time or flight time is

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           t_total = 2 v₀ sin θ / g

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