Question in description

Question

The time it takes for a clock’s pendulum to swing from one side to the other and back again, as shown in the back view of the clock, is called the pendulum’s period. The table shows the period t (in seconds) of a pendulum of length L (in feet).

THE TABLE THAT IS GIVEN

Period, t (sec): 1,2,3,4,5

Length L (ft): 0.82, 3.28, 7.38, 13.12, 20.5

(a) Tell whether the data can be modeled by a linear function, an exponential function, or a quadratic function. Then write an equation for the function.

(b) Find the length of the pendulum that has a period of 0.5 second.

(c) How does decreasing the length of the pendulum by 50% change the period? Justify your answer using several examples.

Answer 1

(a) Since the time is going up by 1 each time (x-value), this makes it a little easier to see if the function is linear, exponential, or quadratic. To find out if the function is linear, you would find out if you have a common difference (or slope). For example, if you take 20.5 - 13.12 = 7.38. If you take 13.12 - 7.38 = 5.74. Since 7.38 ≠ 5.74, this situation is not linear. To find out if the function is exponential, you must have a common ratio, meaning you must multiply (or divide) by the same amount each time to get to the next y-value. So, let's take 20.5/13.12 = 1.5625. Now let's take 13.12/7.38 = 1.77... Since 1.5625 ≠ 1.77..., this function is NOT exponential. To find out if the function is quadratic, the 2nd tier must be added or subtracted by the same amount. So 20.5 - 13.12 = 7.38 and 13.12 - 7.38 = 5.74 and 7.38 - 3.28 = 4.1 and 3.28 - .82 = 2.46. Now let's see if the 2nd tier all add or subtract by the same amount. So let's take 7.38 - 5.74 = 1.64; 5.74 - 4.1 = 1.64; 4.1 - 2.46 = 1.64. Since the 2nd tier have a common difference of 1.64, this would make this situation a QUADRATIC.

(b) The equation for the vertex of a quadratic function is y = a(x - h)² + k, where (h, k) is the vertex. If the pendulum was not moving, it would be vertical and would have no pendulum length. With this in mind, we are going to say the vertex is (0, 0). Using one of the points above (1, 0.82) and a vertex of (0, 0), let's plug these values into the vertex equation and find the "a" value, which is the vertical stretch/shrink of the graph. So 0.82 = a((1) - 0)² + (0). Solving for "a" would get you 0.82. So the equation for this situation is y = 0.82(x - 0)² + 0. So now plug 0.5 into x, and solve for y (pendulum length). So y = 0.82(0.5 - 0)² + 0. So y = 0.205. In other words, at 0.5 seconds, the pendulum length is 0.205 feet.

(c) If the pendulum length was cut in half, then now the lengths of the y-values would be .41, 1.64, 3.69, 6.56, and 10.25. Let's see how this affects the period. So once again, use y = a(x - h)² + k. This means that the "a" value is also cut in half to 0.41. So the new equation is y = 0.41(x - 0)² + 0. So let's plug in a y-value from the list above. Let's use y = 0.41, for example and solve for x to see if anything changes. So 0.41 = 0.41(x)² (we can ignore the (0, 0)). If we solve for x, we get x = 1. So nothing changes here, just that the lengths got shorter.

I hope this helps!