/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 When the length \(L\) of a clock... [FREE SOLUTION] | 91Ó°ÊÓ

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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 \(g=980 \mathrm{cm} / \mathrm{s}^{2}\) to a new location. This increases the period by \(d T=0.001\) s. Find \(d g\) and estimate the value of \(g\) at the new location.

Short Answer

Expert verified
(b) T decreases, clock speeds up. (c) dg = -0.978 cm/s², new g = 979.022 cm/s².

Step by step solution

01

Understand the Relationship between T, L, and g

The period of a pendulum, given by the formula \( T = 2\pi \sqrt{\frac{L}{g}} \), shows how the period \( T \) depends on both the gravitational acceleration \( g \) and the length \( L \) of the pendulum. Here, \( L \) is held constant, and we're analyzing how changes in \( g \) affect \( T \).
02

Differentiate the Period Formula

To find how a small change in \( g \) affects \( T \), we differentiate the formula for \( T = 2\pi \sqrt{\frac{L}{g}} \) with respect to \( g \). Using the chain rule:\[ dT = \frac{d}{dg} \left( 2\pi \sqrt{\frac{L}{g}} \right) \times dg = -\pi \frac{L^{1/2}}{g^{3/2}} \times dg\]
03

Analyze the Effect of Increasing g on T

From the derivative, \( dT = -\pi \frac{L^{1/2}}{g^{3/2}} \times dg \), it is clear that if \( dg \) is positive (i.e., if \( g \) increases), then \( dT \) becomes negative, indicating that \( T \) decreases. This means that a pendulum clock will speed up, since the period, or time taken for each swing, is reduced.
04

Calculate dg using the Given Information

Given: \( L = 100 \) cm and \( g = 980 \) cm/s², with \( dT = 0.001 \) s. We start by using the differentiated relationship:\[ 0.001 = -\pi \frac{100^{1/2}}{980^{3/2}} \times dg\]Solving for \( dg \), we get:\[ dg = -\frac{0.001 \times 980^{3/2}}{\pi \times 10}\]
05

Solve for dg and New g

Calculate \( 980^{3/2} \) and substitute the value back to solve for \( dg \), then compute the new value of \( g \).\[ 980^{3/2} \approx 30687.31 \]\[ dg = -\frac{0.001 \times 30687.31}{3.14159 \times 10} \approx -0.978\]Finally, the new \( g \) is approximately:\[ 980 + (-0.978) = 979.022 \] cm/s².

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Key Concepts

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

Differentiation in Calculus
Differentiation in calculus is all about finding how a function changes when its variables change. It helps us understand the relationships between different factors in an equation. In our pendulum problem, the function we're interested in is the period of the pendulum, represented by the equation:
\[ T = 2\pi \sqrt{\frac{L}{g}} \].
Here, we need to figure out how a small change in the acceleration due to gravity \( g \) results in a change in the period \( T \) of the pendulum. By using differentiation, particularly the chain rule, we can write the derivative of \( T \) with respect to \( g \). This is represented as:
\[ dT = -\pi \frac{L^{1/2}}{g^{3/2}} \times dg \].
This equation helps us to determine \( dT \), a small change in the period, when \( dg \), a small change in gravity, is known. It shows that as \( g \) increases, \( T \) decreases, making the pendulum swing faster and the period shorter.
Acceleration Due to Gravity
The acceleration due to gravity, often referred to as \( g \), is the acceleration of an object caused by Earth's gravitational pull. It's a vital aspect in pendulum clock dynamics, as it directly affects the pendulum's swinging motion.
On Earth, \( g \) varies slightly depending on where you are because the Earth's shape and density are not perfectly uniform. This means that pendulum clocks might speed up or slow down based on their location.
For a pendulum clock, the period \( T \) is directly influenced by \( g \), as seen in the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \].
Thus, if the gravitational acceleration \( g \) increases, the period decreases, causing the pendulum to swing faster. In our example, starting with a known \( g = 980 \) cm/s², we calculated a change, \( dg \), to estimate a new \( g \), showing how sensitive pendulum periods are to changes in \( g \).
Pendulum Clock Dynamics
Pendulum clocks rely on the regular motion of a swinging pendulum to keep accurate time. The core principle behind their operation is the predictability of the pendulum's period \( T \), which depends on both the pendulum's length \( L \) and the local gravitational acceleration \( g \).
The dynamic nature of a pendulum clock means any change in \( g \) can alter the swing timing. This relationship is captured by the formula:
\[ T = 2\pi \sqrt{\frac{L}{g}} \].
As a pendulum clock is relocated, the changes in \( g \) can affect the clock's speed. From the differentiated equation:
\[ dT = -\pi \frac{L^{1/2}}{g^{3/2}} \times dg \],
we know that if \( g \) increases, \( T \) decreases, speeding up the clock. Understanding these dynamics is crucial for calibrating pendulum clocks to new locations to maintain their accuracy. In recent calculations, a pendulum of length 100 cm was moved, increasing its period, allowing us to estimate the local \( g \) and adjust the clock accordingly.

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