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Chapter 3: Problem 39

The period of a pendulum varies directly as the square root of the length of the pendulum and inversely as the square root of the acceleration due to gravity. Find the period when the length is \(121 \mathrm{cm}\) and the acceleration due to gravity is \(980 \mathrm{cm}\) per second squared, if the period is \(6 \pi\) seconds when the length is \(289 \mathrm{cm}\) and the acceleration due to gravity is \(980 \mathrm{cm}\) per second squared.

Short Answer

Expert verified
The period is approximately 3.88468\pi seconds.

Step by step solution

01

Understand the relationship

The period of a pendulum (T) is given by the relationship: \[ T = k \frac{\sqrt{L}}{\sqrt{g}} \] where L is the length of the pendulum, g is the acceleration due to gravity, and k is a constant.
02

Determine the constant (k)

Use the given information to find the value of the constant k. We know that when L = 289 cm and g = 980 cm/s虏, T = 6\pi seconds. Substitute these values into the formula: \[ 6\pi = k \frac{\sqrt{289}}{\sqrt{980}} \] Simplify and solve for k: \[ 6\pi = k \frac{17}{\sqrt{980}} \] \[ k = 6\pi \cdot \frac{\sqrt{980}}{17} \]
03

Simplify the constant (k)

Simplify the expression for k: \[ k = 6\pi \cdot \frac{\sqrt{980}}{17} \approx 6\pi \cdot \frac{31.304951}{17} \approx 6\pi \cdot 1.8411735 \approx 6 \cdot 1.8411735 \pi \approx 11.047041\pi \]
04

Use the constant to find the new period (T)

Now, use the constant k to find the period when L = 121 cm and g = 980 cm/s虏. Substitute these values into the formula: \[ T = 11.047041\pi \cdot \frac{\sqrt{121}}{\sqrt{980}} \] Simplify and solve for T: \[ T = 11.047041\pi \cdot \frac{11}{\sqrt{980}} \] \[ T = 11.047041\pi \cdot \frac{11}{31.304951} \] \[ T = 11.047041\pi \cdot 0.3518800507 \approx 3.88468\pi \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Direct Variation
In mathematics, direct variation describes a linear relationship between two variables where one variable is a constant multiple of the other. For the period of a pendulum, it varies directly as the square root of its length. This means if the length increases, the period of the pendulum also increases proportionally to the square root of the length.
In our example, if we denote the constant of proportionality by k, we can express the direct variation between the period (T) and the length (L) as: \[T \propto \sqrt{L}\] Here, \[\sqrt{L}\] represents the square root of the length of the pendulum. Understanding this relationship helps in calculating how changes in the length will affect the period.
Inverse Variation
Inverse variation is another key concept in understanding pendulums. It involves a relationship between two variables where one variable increases as the other decreases. The period of a pendulum varies inversely as the square root of the acceleration due to gravity (g). This relationship can be written as: \[T \propto \frac{1}{\sqrt{g}} \] This mathematical expression means that when the acceleration due to gravity increases, the period of the pendulum decreases proportionally, and vice versa. In simpler terms, if gravity is stronger (larger g), the pendulum will swing faster (shorter T). Combining this with direct variation, we get the complete formula: \[T = k \frac{\sqrt{L}}{\sqrt{g}}\]
Physics Formulas
Physics often uses formulas to describe relationships between different physical quantities. The formula for the period of a pendulum incorporates both direct and inverse variation to show a comprehensive relationship: \[T = k \frac{\sqrt{L}}{\sqrt{g}} \] Here鈥檚 a quick breakdown of how this works in our example:
  • Calculate the constant (k) using known values: \[6\pi = k \frac{17}{\sqrt{980}} \]. Solve for k to find k = \[11.047041\pi\].
  • Use the value of k to find the period for different lengths and gravitational accelerations. For L = 121 cm and g = 980 cm/s虏:

Substitute these values into the formula: \[T = 11.047041\pi \frac{\sqrt{121}}{\sqrt{980}} \]
Simplify to find the period T: \[T \approx 3.88468\pi \]. Understanding and applying these formulas can help solve various physics problems efficiently.

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Most popular questions from this chapter

Solve each problem. In electric current flow, it is found that the resistance offered by a fixed length of wire of a given material varies inversely as the square of the diameter of the wire. If a wire 0.01 in. in diameter has a resistance of 0.4 ohm, what is the resistance of a wire of the same length and material with diameter 0.03 in., to the nearest ten-thousandth of an ohm?

After the numerator is divided by the denominator, $$f(x)=\frac{x^{5}+x^{4}+x^{2}+1}{x^{4}+1} \quad \text { becomes } \quad f(x)=x+1+\frac{x^{2}-x}{x^{4}+1}$$ (a) What is the oblique asymptote of the graph of the function? (b) Where does the graph of the function intersect its asymptote? (c) As \(x \rightarrow \infty,\) does the graph of the function approach its asymptote from above or below?

Solve each problem. Sum and Product of Two Numbers Find two numbers whose sum is 20 and whose product is the maximum possible value. (Hint: Let \(x\) be one number. Then \(20-x\) is the other number. Form a quadratic function by multiplying them, and then find the maximum value of the function.

Solve each problem. Spending on Shoes and Clothing The total amount spent by Americans on shoes and clothing from 2000 to 2009 can be modeled by $$ f(x)=-4.979 x^{2}+71.73 x+97.29 $$ where \(x=0\) represents 2000 and \(f(x)\) is in billions of dollars. Based on this model, in what year did spending on shoes and clothing reach a maximum? (Source: Bureau of Economic Analysis.)

The number of long-distance phone calls between two cities in a certain time period varies directly as the populations \(p_{1}\) and \(p_{2}\) of the cities and inversely as the distance between them. If \(10,000\) calls are made between two cities \(500 \mathrm{mi}\) apart, having populations of \(50,000\) and \(125,000,\) find the number of calls between two cities 800 mi apart, having populations of \(20,000\) and \(80,000\).
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