Problem 17 - Algebra and Trigonometry with Analytic Geometry

Question

The period \(P\) of a simple pendulum that is, the time required for one complete oscillation-is directly proportional to the square root of its length \(l\). (a) Express \(P\) in terms of \(l\) and a constant of proportionality \(k\). (b) If a pendulum 2 feet long has a period of \(1.5\) seconds, find the value of \(k\) in part (a). (c) Find the period of a pendulum 6 feet long.

Step-by-Step Solution

Answer

(a) \( P = k \sqrt{l} \); (b) \( k \approx 1.06066 \); (c) \( P \approx 2.596 \) seconds.

Step 1: Understanding Direct Proportionality

When a quantity is directly proportional to another, it means it can be expressed as a constant times the other quantity. Here, the period \( P \) of the pendulum is directly proportional to the square root of its length \( l \). Thus, we can write \( P = k \sqrt{l} \), where \( k \) is the constant of proportionality.

Step 2: Finding the Constant of Proportionality

We need to find the value of \( k \) using given information. We are told that when \( l = 2 \) feet, \( P = 1.5 \) seconds. Substitute these values into the equation \( P = k \sqrt{l} \), giving us \( 1.5 = k \sqrt{2} \). Solve for \( k \) by dividing both sides by \( \sqrt{2} \): \( k = \frac{1.5}{\sqrt{2}} \).

Step 3: Calculate \( k \) Numerically

Calculate \( k = \frac{1.5}{\sqrt{2}} \). First, approximate \( \sqrt{2} \approx 1.414 \). Then, divide: \( k \approx \frac{1.5}{1.414} \approx 1.06066 \).

Step 4: Finding the Period of a 6 Feet Pendulum

Now use the value of \( k \) to find the period for a pendulum of length 6 feet. Substitute \( l = 6 \) into the equation \( P = k \sqrt{l} \) and our calculated \( k \approx 1.06066 \). Plug in the values: \( P = 1.06066 \times \sqrt{6} \).

Step 5: Calculate the Period Numerically

First, find \( \sqrt{6} \approx 2.4495 \). Now multiply: \( P = 1.06066 \times 2.4495 \approx 2.596 \) seconds.

Key Concepts

Direct Proportionality

When we say that two quantities are in direct proportionality, it means one quantity increases or decreases exactly in proportion to the other. If Quantity A is directly proportional to Quantity B, we can write this relationship mathematically as \( A = kB \), where \( k \) is a constant known as the constant of proportionality. In our pendulum problem, the period \( P \) is directly proportional to the square root of its length \( l \), expressed as \( P = k \sqrt{l} \). This relationship tells us that as the length \( l \) grows, \( P \) will also increase in a predictable, linear manner. It forms a clear and straightforward way to understand how one variable reacts to the change in another, especially useful in cases like the pendulum's oscillation.

Constant of Proportionality

The constant of proportionality \( k \) helps us understand the specific rate of change in a directly proportional relationship. In simple terms, it determines how "strong" the effect of the variable on the left (\( P \)) is when the variable on the right (\( \sqrt{l} \)) changes. To find \( k \) in the context of the pendulum, we used the given values \( l = 2 \, \text{feet} \) and \( P = 1.5 \, \text{seconds} \). Substituting these into the equation \( 1.5 = k \sqrt{2} \), we solve for \( k \) as \( k = \frac{1.5}{\sqrt{2}} \). This value of \( k \) acts as a bridge to measure the change in the period of the pendulum concerning its length.

Square Root

The square root is a fundamental mathematical operation that finds a number which, when multiplied by itself, equals the original number. In our equation \( P = k \sqrt{l} \), the square root of the length \( l \) is the quantity directly affecting \( P \). For example, if the length of the pendulum is 6 feet, then \( \sqrt{6} \approx 2.4495 \). This process of 'rooting' helps in bringing down measurements to manageable scales or understanding variations over non-linear relationship scales, thus being crucial in physics and engineering problems like ours.

Numerical Approximation

When dealing with irrational numbers like square roots, numerical approximation becomes crucial for practical calculations. We cannot always use exact values such as \( \sqrt{2} \), which is approximately \( 1.414 \). For ease, we approximate to a degree of precision good enough for practical uses without significant loss of detail. In our pendulum scenario, we calculated \( k \approx 1.06066 \) and the period for a 6-foot pendulum as \( P = 1.06066 \times 2.4495 \approx 2.596 \) seconds. Using approximations allows us to solve real-world problems efficiently when exact calculations are computationally intense or unwieldy.