Problem 64 - University Calculus: Early Transcendentals

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 \(100-\mathrm{cm}\) pendulum is moved from a location where \(g=980 \mathrm{cm} / \mathrm{sec}^{2}\) to a new location. This increases the period by \(d T=0.001 \mathrm{sec} .\) Find \(d g\) and estimate the value of \(g\) at the new location.

Step-by-Step Solution

Answer

As \( g \) increases, \( T \) decreases, speeding up the clock. The new \( g \) value is approximately 973.79 cm/s².

Step 1: Differentiate the period equation

Given the equation for the period of the pendulum: \[ T = 2\pi \left(\frac{L}{g}\right)^{1/2} \]We'll differentiate this equation with respect to \( g \) while treating \( L \) as a constant. Using the chain rule, we find:\[ \frac{dT}{dg} = \frac{d}{dg} \left[ 2\pi \left(\frac{L}{g}\right)^{1/2} \right] \]This becomes:\[ \frac{dT}{dg} = 2\pi \cdot \frac{-1}{2} \cdot \left(\frac{L}{g}\right)^{-1/2} \cdot \frac{L}{g^2} \]Simplifying, we obtain:\[ \frac{dT}{dg} = -\pi \left(\frac{L}{g}\right)^{-1/2} \cdot \frac{L}{g^2} \]

Step 2: Substitute values for the initial environment

The length of the pendulum is \( L = 100 \) cm. Initially, the acceleration due to gravity is \( g = 980 \) cm/s². Substitute these values into the simplified expression for \( \frac{dT}{dg} \):\[ \frac{dT}{dg} = -\pi \cdot \left(\frac{100}{980}\right)^{-1/2} \cdot \frac{100}{980^2} \]

Step 3: Calculate the derivative

Next, compute the numerical approximation of the derivative:\[ \frac{dT}{dg} = -\pi \cdot \left(\frac{100}{980}\right)^{-1/2} \cdot \frac{100}{980^2} \approx -1.61 \times 10^{-4} \text{ sec/cm} \]

Step 4: Determine the effect of increased g

To answer part (b), note that since \( \frac{dT}{dg} < 0 \), as \( g \) increases, \( T \) decreases. This means that a pendulum clock will speed up because it takes less time to complete one cycle.

Step 5: Solve for dg in the new environment

Given \( dT = 0.001 \) sec, we can use the relationship \( dT = \frac{dT}{dg} \cdot dg \) to find \( dg \).\[ dg = \frac{dT}{\frac{dT}{dg}} = \frac{0.001}{-1.61 \times 10^{-4}} \approx -6.21 \text{ cm/s}^2 \]

Step 6: Calculate new value of g

Finally, estimate the new value of \( g \) by subtracting \( dg \) from the original \( g \) value:\[ g_{\text{new}} = 980 + (-6.21) = 973.79 \text{ cm/s}^2 \]

Key Concepts

Acceleration of Gravity

The acceleration of gravity, denoted by the symbol \( g \), is the force that gives objects their weight and causes them to fall towards the Earth. It varies slightly depending on location due to Earth's shape and rotation.
For example, gravity is often stronger at the poles compared to the equator. That's because the Earth is not a perfect sphere. It has a slight bulge at the equator.

• This variation can affect precise measurements that rely on time and distance, like the swing of a pendulum.
  
• In pendulum motion, understanding how \( g \) affects the period \( T \) of the pendulum is essential. It helps us understand how clocks can be slightly off if moved to different geographical locations.
  

Changes in \( g \) can lead to shifts in the period, thereby directly impacting the accuracy of pendulum clocks.

Period of a Pendulum

The period of a pendulum, \( T \), is the time it takes for the pendulum to complete one full swing back and forth. For a simple pendulum, this period is determined by the length of the pendulum \( L \) and the local acceleration of gravity \( g \). The equation describing this is:
\[ T = 2\pi \left( \frac{L}{g} \right)^{1/2} \]
Here's how each component affects the period:

• **Length \( L \):** The longer the pendulum, the longer the period. This means it takes more time for the pendulum to swing because it covers a greater distance.
  
• **Gravity \( g \):** If gravity increases, the period decreases. The pendulum takes less time to complete a swing because the stronger gravitational pull provides a quicker acceleration.
  

Understanding this relationship is crucial for accurate timekeeping and in scientific studies where measuring \( g \) accurately is important.

Differentiation

Differentiation is a mathematical process to find the rate at which one quantity changes with respect to another. In the context of pendulum motion, differentiation helps us determine how a change in gravity \( g \) affects the period \( T \) of the pendulum.
To differentiate the period equation \[ T = 2\pi \left( \frac{L}{g} \right)^{1/2} \] with respect to \( g \), we apply the chain rule. Here's a simplified breakdown:

• The derivative \( \frac{dT}{dg} \) represents how small changes in \( g \) lead to changes in \( T \).
  
• Since \( \frac{dT}{dg} \) is negative, an increase in \( g \) results in a decrease in \( T \), meaning the clock speeds up.
  

This is critical for calculating shifts in period caused by changes in gravity if a pendulum clock is moved to a location with a different \( g \). Differentiation thus provides a deeper understanding of the specific relationship between gravity and the time-measuring properties of a pendulum.