/*! 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 Measuring acceleration of gravit... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 54

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 \(L .\) $$ \begin{array}{l}{\text { a. With } L \text { held constant and } g \text { as the independent variable, }} \\ {\text { calculate } d T \text { and use it to answer parts (b) and (c). }} \\ {\text { b. If } g \text { increases, will } T \text { increase or decrease? Will a pendulum }} \\ {\text { clock speed up or slow down? Explain. }}\end{array} $$ $$ \begin{array}{l}{\text { c. A clock with a } 100 \text { -cm pendulum is moved from a location }} \\ {\text { where } g=980 \mathrm{cm} / \mathrm{sec}^{2} \text { to a new location. This increases the }} \\ {\text { period by } d T=0.001 \mathrm{sec} . \text { Find } d g \text { and estimate the value of }} \\\ {g \text { at the new location. }}\end{array} $$

Short Answer

Expert verified
a) \( dT = -\pi\frac{L^{1/2}}{g^{3/2}} \cdot dg \). b) T decreases if g increases, so the clock speeds up. c) \( g_{new} \approx 979.99 \text{ cm/s}^2 \).

Step by step solution

01

Differentiate the period formula

We start with the formula for the pendulum's period: \( T = 2\pi\sqrt{\frac{L}{g}} \). To find the differential change in \( T \), denoted as \( dT \), with \( g \) as the independent variable, we differentiate the period formula with respect to \( g \). This gives:\[dT = \frac{d}{dg}\left(2\pi\sqrt{\frac{L}{g}}\right) \cdot dg\]We use the chain rule to differentiate:\[\frac{d}{dg}\left(2\pi\sqrt{\frac{L}{g}}\right) = -\pi\frac{L^{1/2}}{g^{3/2}}\]Therefore:\[ dT = -\pi\frac{L^{1/2}}{g^{3/2}} \cdot dg \]
02

Analyze how changes in gravity affect period

From the expression \( dT = -\pi\frac{L^{1/2}}{g^{3/2}} \cdot dg \), we see that if \( g \) increases (i.e., \( dg > 0 \)), then \( dT < 0 \). This means that the period \( T \) will decrease when \( g \) increases. Consequently, the pendulum clock will run faster (speed up) since its period is shorter.
03

Calculate \( dg \) and new \( g \) value

Given that \( L = 100 \) cm and originally \( g = 980 \) cm/s虏, we need to find \( dg \) given \( dT = 0.001 \) seconds. From the previous step's result:\[ \begin{align*}0.001 &= -\pi\frac{100^{1/2}}{980^{3/2}} \cdot dg \dg &= \frac{0.001}{-\pi\frac{10}{980^{3/2}}}\end{align*} \]Calculate \( dg \):\[ dg \approx -10.11 \times 10^{-3}\]The new value of \( g \) is:\[ g_{new} = 980 + dg = 980 - 0.01011 \approx 979.98989 \text{ cm/s}^2 \]

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus, used to determine how functions change. In this context, we're focusing on how the period of a pendulum, denoted as \( T \), changes in response to variations in gravitational acceleration, \( g \). For a pendulum, the period \( T \) depends on \( g \) according to the formula: \\( T = 2\pi\sqrt{\frac{L}{g}} \). Here, \( L \) is the constant length of the pendulum. To find out how a small change in \( g \) affects \( T \), we calculate the differential \( dT \). This involves differentiating the function with respect to \( g \). \
  • The differentiation process involves the chain rule, which helps us handle composite functions.
In this specific computation, we find: \\[\frac{d}{dg}\left(2\pi\sqrt{\frac{L}{g}}\right) = -\pi\frac{L^{1/2}}{g^{3/2}}\] \By multiplying this result by \( dg \), the small change in \( g \), we get \( dT \). Differentiating provides insight into the impact that minor variations in gravity have on the pendulum's period.
Gravity
Gravity is the force that attracts two masses towards each other. On Earth, it gives weight to objects and influences various physical phenomena. With pendulums, gravity impacts the period \( T \), the time it takes for one complete oscillation. \
  • The force of gravity affects not only the motion of pendulums but also systems like tides and astronomical bodies.
  • Its strength varies slightly from location to location on the Earth's surface due to factors like altitude and Earth's shape.
\In this problem, we see how a change in gravity changes the period of a pendulum. Mathematically, it is represented in our differentiated formula: \( dT = -\pi\frac{L^{1/2}}{g^{3/2}} \cdot dg \). This shows that an increase in gravity (\( dg > 0 \)) decreases the period \( T \), as the expression is negative. Lesser period means the pendulum completes its swing faster. Understanding gravity's role helps in grasping how precise instruments like pendulum clocks must adjust to slight changes for accurate timekeeping.
Acceleration
Acceleration, in physics, is the rate of change of velocity of an object. When discussing a pendulum, we usually talk about gravitational acceleration, which is the acceleration experienced due to Earth's gravity. In the pendulum formula \( T = 2\pi\sqrt{\frac{L}{g}} \), \( g \) is the gravitational acceleration. This acceleration influences how fast or slow the pendulum moves through its path. \
  • A key aspect of gravitational acceleration is that it can vary slightly depending on where you are on Earth, although it's generally around \( 9.8 \text{ m/s}^2 \) (or \( 980 \text{ cm/s}^2 \)).
  • In this exercise, acceleration was indirectly measured through how it affects the pendulum's period.
\When \( g \) increases, because of the inverse relationship with \( \sqrt{g} \) in the period formula, \( T \) decreases. We calculated \( dg \), the change in gravitational acceleration using \( dT = 0.001 \text{ sec} \). Solving for \( dg \), we find the shift necessary in \( g \) to result in such a period change, highlighting how sensitive pendulum timekeeping is to acceleration caused by gravity.

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