Problem 52 - Thomas Calculus

Question

Measuring acceleration of gravity When the length \(L\) of a clock pendulum is held constant by controlling its temperature, the pendulum's period \(T\) depends on the acceleration of gravity \(g .\) The period will therefore vary slightly as the clock is moved from place to place on the earth's surface, depending on the change in \(g .\) By keeping track of \(\Delta T,\) we can estimate the variation in \(g\) from the equation \(T=2 \pi(L / g)^{1 / 2}\) that relates \(T, g,\) and \(L\) a. With \(L\) held constant and \(g\) as the independent variable, calculate \(d T\) and use it to answer parts \((b)\) and \((c)\) . b. If \(g\) increases, will \(T\) increase or decrease? Will a pendulum clock speed up or slow down? Explain. c. A clock with a 100 -cm pendulum is moved from a location were \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001\) sec. Find \(d g\) and estimate the value of \(g\) at the new location.

Step-by-Step Solution

Answer

T decreases with increased g; pendulum clock speeds up. New g is approximately 979.9024 cm/sec².

Step 1: Differentiate Period Equation with Respect to g

The period equation is given as \( T = 2\pi \sqrt{\frac{L}{g}} \). To find \( dT \), we differentiate with respect to \( g \) while keeping \( L \) constant.\[ \frac{dT}{dg} = \frac{d}{dg}\left( 2\pi \sqrt{\frac{L}{g}} \right) \]Applying the chain rule, we have:\[ \frac{dT}{dg} = 2\pi \cdot \frac{1}{2}\cdot \left(\frac{L}{g}\right)^{-1/2} \cdot \left( -\frac{L}{g^2} \right) \]\[ \frac{dT}{dg} = -\pi \cdot \frac{L}{g^{3/2}} \]

Step 2: Determine Change in Period with Change in Gravity (b)

Since \( \frac{dT}{dg} = -\pi \cdot \frac{L}{g^{3/2}} \), when \( g \) increases, \( \frac{dT}{dg} \) is negative, indicating that \( T \) decreases. Therefore, the period \( T \) of the pendulum decreases, causing the pendulum clock to speed up.

Step 3: Calculate Change in Gravity (c)

Given: \( L = 100 \text{ cm} \), initial \( g = 980 \text{ cm/sec}^2 \), and change in period \( dT = 0.001 \text{ sec} \).Using the derivative from Step 1:\[ dT = \frac{dT}{dg} \cdot dg = -\pi \cdot \frac{L}{g^{3/2}} \cdot dg \]Solving for \( dg \):\[ dg = \frac{dT}{-\pi \cdot \frac{L}{g^{3/2}}} \]Substitute \( L = 100 \), \( g = 980 \), and \( dT = 0.001 \):\[ dg = \frac{0.001}{-\pi \cdot \frac{100}{980^{3/2}}} \]Calculate \( dg \):\[ \text{First calculate } g^{3/2}: 980^{3/2} \approx 30625.485 \ \frac{100}{30625.485} \approx 0.00326 \]\[ dg \approx \frac{0.001}{-\pi \cdot 0.00326} \approx \frac{0.001}{-0.01024} \approx -0.0976 \]Thus, the change in gravity is approximately \( -0.0976 \text{ cm/sec}^2 \).

Step 4: Estimate New Gravity

To find the new value of \( g \), subtract \( dg \) from the original value:\[ g_{\text{new}} = g_{\text{old}} + dg = 980 - 0.0976 \approx 979.9024 \text{ cm/sec}^2 \]

Key Concepts

Acceleration of Gravity

The acceleration due to gravity, commonly denoted as \( g \), is a fundamental concept that describes the force pulling objects towards the Earth's surface. Its value can vary slightly depending on where you are on Earth. Typically, \( g \) is approximately 9.81 m/s²; however, it's often measured as 980 cm/s² for precision in exercises.

• In the context of a pendulum, the acceleration of gravity affects the pendulum's period (the time it takes for one full swing back and forth).
• This period is mathematically represented by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( L \) is the length of the pendulum.

As \( g \) changes, so does the period \( T \). Understanding how tiny changes in \( g \) can affect pendulum movement is crucial for precise timekeeping in pendulum clocks. This makes \( g \) an essential factor when moving clocks to different locations.

Differentiation

In mathematics, differentiation is a way of finding how a function changes as its input changes. For a pendulum, the period \( T \) is a function of gravity \( g \). Differentiation helps us understand how small changes in \( g \) affect \( T \).
To differentiate \( T = 2\pi \sqrt{\frac{L}{g}} \) with respect to \( g \), we apply this concept to find \( \frac{dT}{dg} \), which represents the rate of change of the period with respect to gravity.
This differentiation gives us: \[ \frac{dT}{dg} = -\pi \cdot \frac{L}{g^{3/2}} \]

• The negative sign tells us that as gravity increases, the pendulum's period decreases.
• This formula is derived by applying the chain rule of differentiation.

Such differentiation is essential for making precise adjustments to pendulum clocks when they are moved to areas with different gravitational acceleration values.

Chain Rule

The chain rule is a fundamental technique used in calculus to differentiate composite functions. When dealing with pendulum periods, where \( T = 2\pi \sqrt{\frac{L}{g}} \), the formula involves a square root, which is a function within a function. Here, the chain rule becomes vital.
The chain rule states that if you have a composite function \( f(g(x)) \), its derivative is \( f'(g(x)) \cdot g'(x) \).
For our pendulum scenario:

• First, we consider the outer function \( 2\pi \sqrt{u} \), where \( u = \frac{L}{g} \).
• Its derivative with respect to \( u \) is \( \frac{1}{2u^{1/2}} \).
• Next, we differentiate \( u \) with respect to \( g \) to obtain \( -\frac{L}{g^2} \).

Multiply these together to find \( \frac{dT}{dg} \), integrating both parts through the chain rule, combining them:\[ \frac{dT}{dg} = 2\pi \cdot \frac{1}{2} \cdot \left(\frac{L}{g}\right)^{-1/2} \cdot \left(-\frac{L}{g^2}\right) = -\pi \cdot \frac{L}{g^{3/2}} \]
This process shows how interconnected changes, such as adjustments in gravity, flow through the function.

Pendulum Clock

A pendulum clock uses a swinging pendulum to mark the passage of time. The period of the pendulum, directly impacted by gravity, determines the clock's accuracy. However, moving the clock to a location with a different gravitational force alters its timekeeping.
In exercises, understanding how \( g \) affects the period \( T \) allows us to estimate the adjustment required for accurate time. For example:

• If gravity increases, \( T \) decreases, causing the clock to run faster.
• If gravity decreases, the opposite happens, and \( T \) increases, causing the clock to slow down.

These changes are crucial when a pendulum clock is moved between different geographic locations. In practical applications, such as the mentioned exercise, calculating \( dT \) when \( g \) changes helps keep time precisely despite the change in location. This emphasizes the importance of understanding gravitational variations when working with or relocating pendulum clocks.