/*! 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 88 A grandfather clock has a simple... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A grandfather clock has a simple brass pendulum of length L. One night, the temperature in the house is \(25.0^{\circ} \mathrm{C}\) and the period of the pendulum is 1.00 s. The clock keeps correct time at this temperature. If the temperature in the house quickly drops to \(17.1^{\circ} \mathrm{C}\) just after 10 P.M., and stays at that value, what is the actual time when the clock indicates that it is 10 A.M. the next morning?

Short Answer

Expert verified
The actual time when the clock indicates it is 10 A.M. is after 10 A.M.

Step by step solution

01

Understanding the Relationship

The period of a simple pendulum is given by \( T = 2\pi \sqrt{\frac{L}{g}} \), where \( T \) is the period, \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity. Here, the period at \( 25.0^{\circ} \mathrm{C} \) is correctly 1.00 s.
02

Find Change in Length

The length of the pendulum changes with temperature. The change in length \( \Delta L \) can be described by \( \Delta L = \alpha L_0 \Delta T \), where \( \alpha \) is the coefficient of linear expansion for brass \( (1.9 \times 10^{-5} /^{\circ}C) \), \( L_0 \) is the original length, and \( \Delta T \) is the temperature change: \( 17.1^{\circ}C - 25.0^{\circ}C = -7.9^{\circ}C \).
03

Calculate New Length

Substitute the values into \( L + \Delta L = L(1 + \alpha \Delta T) \). Since \( \Delta L = \alpha L_0 \Delta T \), calculate the change as a factor to the original length.
04

New Period Calculation

With the new length, the new period \( T' \) becomes \( T' = 2\pi \sqrt{\frac{L'}{g}} \). Substitute \( L' = L(1 + \alpha \Delta T) \) into the period formula.
05

Determine Time Discrepancy

Calculate how the altered period affects the time. If \( T' > 1.00 \) then the clock is running slow. Multiply \( T' \) by the number of seconds in 12 hours (from 10 PM to 10 AM) to find the real elapsed time.
06

Calculate Actual Time

Determine the actual clock reading by subtracting or adding the discrepancy over 12 hours from the new calculated real elapsed time. This gives when it actually hits 10 A.M.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Temperature Effect on Pendulum
Temperature changes can significantly affect the period of a pendulum. A pendulum’s length is sensitive to temperature, especially if made from materials like brass, which expand or contract with temperature fluctuations.

In the case of a brass pendulum, when the temperature decreases, the pendulum length will become shorter due to contraction. Since the pendulum’s length is directly related to its period, any change in length can also change the period. The formula for the period of a simple pendulum is:
  • \[ T = 2\pi \sqrt{\frac{L}{g}} \]
Where \( T \) is the period, \( L \) is the length, and \( g \) is the acceleration due to gravity.

With a drop in temperature, if the pendulum shortens, it results in a slightly shorter period. Thus, the pendulum will swing faster, causing the clock to run slightly ahead of time. This effect must be considered for accurate timekeeping in grandfather clocks.
Coefficient of Linear Expansion
The coefficient of linear expansion is a measure of how much a material's length changes with a change in temperature. Brass, the material our pendulum is made of, has a coefficient of linear expansion denoted by \( \alpha \), which is typically around \( 1.9 \times 10^{-5} /^{\circ}C \).

To calculate the change in the pendulum's length due to temperature change, we use the formula:
  • \[ \Delta L = \alpha L_0 \Delta T \]
Where \( \Delta L \) is the change in length, \( \alpha \) is the coefficient of linear expansion, \( L_0 \) is the original length, and \( \Delta T \) is the temperature change. For our problem, \( \Delta T \) is \(-7.9^{\circ}C\). This negative change indicates contraction.

Understanding the coefficient of linear expansion is crucial for predicting how temperature will impact materials used in construction, as well as delicate mechanical devices like pendulums.
Simple Harmonic Motion
Pendulums are a classic example of simple harmonic motion (SHM). SHM is characterized by a repetitive back and forth movement through a central equilibrium position, typically resulting in a sine wave motion over time.

In a pendulum, SHM occurs due to the force of gravity acting on the suspended mass, causing it to move through its arc with periodic and predictable motion. The period of a simple pendulum within SHM is primarily influenced by its length but also, as in this case, by external factors like temperature.

As the pendulum lengthens or shortens with temperature adjustments, the period will change, altering how quickly the pendulum swings. A longer pendulum has a longer period, resulting in slower swings, while a shorter pendulum has a shorter period, leading to faster swings.

Understanding SHM is essential as it helps predict and explain the behavior of systems that can oscillate, like pendulums, springs, and even alternating current in electricity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

More than likely, there is a glowing incandescent lightbulb in your room at this moment. The filament of that bulb, with a temperature of about \(4500^{\circ} \mathrm{F},\) is almost half as hot as the surface of the Sun. What is this temperature in degrees Celsius?

When the bulb of a constant-volume gas thermometer is placed in a beaker of boiling water at \(100^{\circ} \mathrm{C}\), the pressure of the gas is \(227 \mathrm{mmHg}\). When the bulb is moved to an ice-salt mixture, the pressure of the gas drops to \(162 \mathrm{mmHg}\). Assuming ideal behavior, as in Figure \(16-3\), what is the Celsius temperature of the ice-salt mixture?

When technicians work on a computer, they often ground themselves to prevent generating a spark. If an electrostatic discharge does occur, it can cause temperatures as high as \(1500^{\circ} \mathrm{C}\) in a localized area of a circuit. Temperatures this high can melt aluminum, copper, and silicon. What is this temperature in (a) degrees Fahrenheit and (b) kelvins?

Heat Transport in the Human Body The core temperature of the human body is \(37.0^{\circ} \mathrm{C}\), and the skin, with a surface area of \(1.40 \mathrm{m}^{2},\) has a temperature of \(34.0^{\circ} \mathrm{C}\). (a) Find the rate of heat transfer out of the body under the following as sumptions: (i) The average thickness of tissue between the core and the skin is \(1.20 \mathrm{cm} ;\) (ii) the thermal conductivity of the tissue is that of water. (b) Without repeating the calculation of part (a), what rate of heat transfer would you expect if the skin temperature were to fall to \(31.0^{\circ} \mathrm{C}\) ? Explain.

Double-Paned Window An energy-efficient doublepaned window consists of two panes of glass, each with thickness \(L_{1}\) and thermal conductivity \(k_{1}\), separated by a layer of air of thickness \(L_{2}\) and thermal conductivity \(k_{2}\). Show that the equilibrium rate of heat flow through this window per unit area, \(A\), is $$\frac{Q}{A t}=\frac{\left(T_{2}-T_{1}\right)}{2 L_{1} / k_{1}+L_{2} / k_{2}} $$In this expression, \(T_{1}\) and \(T_{2}\) are the temperatures on either side of the window.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.