Problem 24
Question
\(\bullet\) Find the period, frequency, and angular frequency of (a) the second hand and (b) the minute hand of a wall clock.
Step-by-Step Solution
Verified Answer
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.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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.
4Step 4: 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.
5Step 5: Determining Frequency of the Minute Hand
For the minute hand, the frequency \( f \) is \( f = \frac{1}{T} = \frac{1}{3600} \) Hz.
6Step 6: 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.
7Step 7: 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.
Key Concepts
Frequency CalculationAngular FrequencyClocks and Timekeeping
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.
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.
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.
Other exercises in this chapter
Problem 21
\(\bullet\) \(\bullet\) A steel cable with cross-sectional area of 3.00 \(\mathrm{cm}^{2}\) has an elastic limit of \(2.40 \times 10^{8}\) Pa. Find the maximum
View solution Problem 23
\(\bullet\) (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern having the same frequency of the note that is sung. If someon
View solution Problem 25
\(\bullet\) If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it oscillates. Suppose it is displaced 0.12
View solution Problem 27
\(\bullet\) The wings of the Blue-throated Hummingbird (Lampornis clemenciae), which inhabits Mexico and the southwestern United States, beat at a rate of up to
View solution