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Chapter 11: Problem 24

\(\bullet\) Find the period, frequency, and angular frequency of (a) the second hand and (b) the minute hand of a wall clock.

Short Answer

Expert verified
The period of the second hand is 60 seconds, the frequency is \( \frac{1}{60} \) Hz, and the angular frequency is \( \frac{\pi}{30} \) rad/s. The period of the minute hand is 3600 seconds, the frequency is \( \frac{1}{3600} \) Hz, and the angular frequency is \( \frac{\pi}{1800} \) rad/s.

Step by step solution

01

Understanding the Concept of Period

The period of a repeating event is the duration of one complete cycle. In the case of a clock, the second hand completes one full cycle in 60 seconds, and the minute hand completes one full cycle in 60 minutes, or 3600 seconds.
02

Calculating the Period of the Second Hand

The second hand of a clock completes a full cycle every 60 seconds. Therefore, its period is 60 seconds.
03

Calculating the Period of the Minute Hand

The minute hand of a clock completes a full cycle every 3600 seconds (60 minutes multiplied by 60 seconds per minute). Therefore, its period is 3600 seconds.
04

Determining Frequency of the Second Hand

The frequency is the reciprocal of the period. For the second hand, frequency \( f \) can be calculated as \( f = \frac{1}{T} = \frac{1}{60} \) Hz.
05

Determining Frequency of the Minute Hand

For the minute hand, the frequency \( f \) is \( f = \frac{1}{T} = \frac{1}{3600} \) Hz.
06

Calculating Angular Frequency of the Second Hand

Angular frequency \( \omega \) is given by the formula \( \omega = 2\pi f \). For the second hand, \( \omega = 2\pi \times \frac{1}{60} \), which simplifies to \( \frac{\pi}{30} \) radians per second.
07

Calculating Angular Frequency of the Minute Hand

For the minute hand, \( \omega = 2\pi \times \frac{1}{3600} \), which simplifies to \( \frac{\pi}{1800} \) radians per second.

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Key Concepts

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

Frequency Calculation
Calculating frequency involves understanding how often a repeating event occurs within a specific timeframe. Frequency is denoted by the symbol \( f \) and is measured in hertz (Hz), where one hertz equates to one cycle per second. In the context of a clock:
  • The second hand completes a full rotation every 60 seconds. Therefore, its frequency is determined using the formula \( f = \frac{1}{T} \), where \( T \) is the period. So, for the second hand, \( f = \frac{1}{60} \) Hz.
  • Similarly, the minute hand takes 3600 seconds to complete a cycle, making its frequency \( f = \frac{1}{3600} \) Hz.
And here's a tip: the larger the period, the smaller the frequency, as they are inversely proportional. Using these principles, you can calculate the frequency of any oscillating system by first determining its period and then applying the simple reciprocal formula.
Angular Frequency
Angular frequency is a concept that describes how fast something rotates or oscillates in relation to a full circle, measured in radians per second. It is given by the formula \( \omega = 2\pi f \), where \( \omega \) is angular frequency and \( f \) is the frequency.
  • For the second hand of a clock, with a frequency of \( \frac{1}{60} \) Hz, the angular frequency becomes \( \omega = 2\pi \times \frac{1}{60} = \frac{\pi}{30} \) radians per second.
  • For the minute hand, \( \omega = 2\pi \times \frac{1}{3600} = \frac{\pi}{1800} \) radians per second.
Angular frequency provides insight into how quickly an object is oscillating in terms of angles, which is particularly useful in systems involving circular motion or harmonic oscillators. It offers a different perspective from linear frequency, emphasizing the rotational aspect of motion.
Clocks and Timekeeping
Clocks are excellent examples of oscillatory systems in everyday life, utilizing consistent oscillations to keep track of time. The second and minute hands exhibit periodic motion, making it easier to calculate not only their period but also their frequency and angular frequency.
  • The second hand's 60-second period indicates a slow but steady pace, allowing us to measure seconds accurately using its known frequency and angular frequency.
  • The minute hand's 3600-second period aids in tracking longer intervals, such as minutes.
Understanding the mechanism of clocks helps in appreciating the accuracy and precision provided by these devices. The regulated motion of the hands, governed by precise periods and frequencies, ensures that time is measured uniformly across various timepieces, aiding everything from global communication to daily scheduling tasks.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) An apple weighs 1.00 \(\mathrm{N}\) . When you hang it from the end of a long spring of force constant 1.50 \(\mathrm{N} / \mathrm{m}\) and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back-and-forth swings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring (with the apple removed)?

\(\bullet\) A proud deep-sea fisherman hangs a 65.0 \(\mathrm{kg}\) fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m} .\) (a) What is the force constant of the spring? (b) What is the period of oscillation of the fish if it is pulled down 3.50 \(\mathrm{cm}\) and released?

\(\bullet\) \(\bullet\) One end of a stretched ideal spring is attached to an airtrack and the other is attached to a glider with a mass of 0.355 \(\mathrm{kg}\) . The glider is released and allowed to oscillate in SHM. If the distance of the glider from the fixed end of the spring varies between 1.80 \(\mathrm{m}\) and \(1.06 \mathrm{m},\) and the period of the oscillation is \(2.15 \mathrm{s},\) find (a) the force constant of the spring, (b) the maximum speed of the glider, and (c) the magnitude of the maximum acceleration of the glider.

\(\bullet\) A 0.150 \(\mathrm{kg}\) toy is undergoing SHM on the end of a horizontal spring with force constant 300.0 \(\mathrm{N} / \mathrm{m} .\) When the object is 0.0120 \(\mathrm{m}\) from its equilibrium position, it is observed to have a speed of 0.300 \(\mathrm{m} / \mathrm{s}\) . Find (a) the total energy of the object at any point in its motion, (b) the amplitude of the motion, and (c) the maximum speed attained by the object during its motion.

\(\bullet\) In the Challenger Deep of the Marianas Trench, the depth of seawater is 10.9 \(\mathrm{km}\) and the pressure is \(1.16 \times 10^{8}\) Pa (about 1150 atmospheres). (a) If a cubic meter of water is taken to this depth from the surface (where the normal atmospheric pressure is about \(1.0 \times 10^{3} \mathrm{Pa}\) , what is the change in its volume? Assume that the bulk modulus for seawater is the same as for freshwater \(\left(2.2 \times 10^{9} \mathrm{Pa}\right) .\) (b) At the surface, seawater has a density of \(1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) What is the density of sea-water at the depth of the Challenger Deep?
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