Problem 37
Question
The period (the time required for one complete oscillation) of a simple pendulum varies directly as the square root of its length. If a pendulum 9 inches long has a period of \(2.4\) seconds, find the period of a pendulum of length 12 inches. Express the answer to the nearest tenth of a second. \(2.8\) seconds
Step-by-Step Solution
Verified Answer
The period is approximately 2.8 seconds.
1Step 1: Understand the Relationship
The problem states that the period \(T\) of a pendulum varies directly as the square root of its length \(L\). This can be expressed as \(T = k\sqrt{L}\), where \(k\) is a constant of proportionality.
2Step 2: Determine the Constant
We start by using the given information: a pendulum 9 inches long has a period of 2.4 seconds. Plug these values into the equation: \(2.4 = k\sqrt{9}\). Simplifying this, we get \(2.4 = 3k\), so \(k = \frac{2.4}{3} = 0.8\).
3Step 3: Apply the Constant to New Length
Now, we need to find the period of a pendulum that is 12 inches long. Using the equation \(T = k\sqrt{L}\) with \(k = 0.8\) and \(L = 12\), compute \(T = 0.8\sqrt{12}\).
4Step 4: Calculate the Period for 12 Inches
Calculate the square root of 12, which is approximately 3.4641. Then, multiply this by the constant: \(T = 0.8 \times 3.4641 \approx 2.7713\).
5Step 5: Round the Period
Round the calculated period to the nearest tenth, giving \(T \approx 2.8\) seconds.
Key Concepts
Direct Variation ExplainedUnderstanding the Square Root FunctionRole of the Proportionality ConstantPendulum Length Impact on Period
Direct Variation Explained
Direct variation is a fundamental concept in mathematics, representing a specific type of relationship between two variables. When two quantities vary directly, it means that when one quantity changes, the other changes in a consistent and proportional manner.
In the context of our pendulum problem, the period of the pendulum (denoted as \(T\)) varies directly with the square root of its length (denoted as \(L\)). This relationship can be expressed mathematically as \(T = k\sqrt{L}\), where \(k\) is the proportionality constant. Knowing this helps us understand how changes in the pendulum's length affect the pendulum's period.
In the context of our pendulum problem, the period of the pendulum (denoted as \(T\)) varies directly with the square root of its length (denoted as \(L\)). This relationship can be expressed mathematically as \(T = k\sqrt{L}\), where \(k\) is the proportionality constant. Knowing this helps us understand how changes in the pendulum's length affect the pendulum's period.
Understanding the Square Root Function
The square root function is an important mathematical concept that is essential in understanding the relationship between the pendulum's length and its period.
In our equation, the pendulum's period \(T\) is proportional to the square root of the length \(L\), represented as \(\sqrt{L}\). Here, the square root operation means finding a number which, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because \(3 \times 3 = 9\).
This function is significant because it showcases how the pendulum’s period doesn't linearly rise with the pendulum's length. Instead, it scales by the 'milder' rate of the square root, representing this less-than-direct escalation.
In our equation, the pendulum's period \(T\) is proportional to the square root of the length \(L\), represented as \(\sqrt{L}\). Here, the square root operation means finding a number which, when multiplied by itself, gives the original number. For example, the square root of 9 is 3 because \(3 \times 3 = 9\).
This function is significant because it showcases how the pendulum’s period doesn't linearly rise with the pendulum's length. Instead, it scales by the 'milder' rate of the square root, representing this less-than-direct escalation.
Role of the Proportionality Constant
The proportionality constant, denoted as \(k\), plays a crucial role in problems involving direct variation.
In our pendulum's period equation \(T = k\sqrt{L}\), \(k\) determines how the period scales with the square root of the length. To find \(k\), we use known values, such as the given length and period of a pendulum, which allows us to calculate this constant for other scenarios.
In our exercise, by plugging in the 9-inch pendulum length and its 2.4-second period, we solve for \(k\) as \(k = \frac{2.4}{3} = 0.8\). With this constant, we easily compute the periods of pendulums of different lengths by maintaining this proportional relationship.
In our pendulum's period equation \(T = k\sqrt{L}\), \(k\) determines how the period scales with the square root of the length. To find \(k\), we use known values, such as the given length and period of a pendulum, which allows us to calculate this constant for other scenarios.
In our exercise, by plugging in the 9-inch pendulum length and its 2.4-second period, we solve for \(k\) as \(k = \frac{2.4}{3} = 0.8\). With this constant, we easily compute the periods of pendulums of different lengths by maintaining this proportional relationship.
Pendulum Length Impact on Period
The length of a pendulum significantly impacts its period, following a predictable pattern due to the direct variation relationship with the square root function.
As we increase the pendulum's length, the time it takes to complete one oscillation grows, but not in a straightforward fashion. Instead, the period increases with the square root of the length. This means that even a sizable increase in length results in a less dramatic increase in the pendulum's period.
As we increase the pendulum's length, the time it takes to complete one oscillation grows, but not in a straightforward fashion. Instead, the period increases with the square root of the length. This means that even a sizable increase in length results in a less dramatic increase in the pendulum's period.
- For example, increasing the length from 9 inches to 12 inches involves calculating their square roots (3 and approximately 3.4641, respectively), showing a less substantial rise.
- The final step involves multiplying this by our constant \(k = 0.8\) to find the specific period, demonstrating how mathematical insights translate to real-world implications.
Other exercises in this chapter
Problem 36
Graph \(f(x)=\sqrt{x^{2}+8}, f(x)=\sqrt{x^{2}+4}\), and \(f(x)=\) \(\sqrt{x^{2}+1}\) on the same set of axes. Look at these graphs and predict the graph of \(f(
View solution Problem 36
Explain the concept of a piecewise-defined function.
View solution Problem 37
For each of the following, predict the general shape and location of the graph, and then use your calculator to graph the function to check your prediction. (a)
View solution Problem 37
Is \(f(x)=\left(3 x^{2}-2\right)-(2 x+1)\) a quadratic function? Explain your answer.
View solution