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Two carpenters are hammering at the same time, each at a different hammering frequency. The hammering frequency is the number of hammer blows per second. Every \(4.6 \mathrm{s},\) both carpenters strike at the same instant, producing an effect very similar to a beat frequency. The first carpenter strikes a blow every 0.75 s. How many seconds elapse between the second carpenter's blows if the second carpenter hammers (a) more rapidly than the first carpenter, and (b) less rapidly than the first carpenter?

Short Answer

Expert verified
(a) 0.65 s; (b) 0.90 s.

Step by step solution

01

Determine the First Carpenter's Hammering Frequency

Since the first carpenter strikes a blow every 0.75 seconds, his hammering frequency is the reciprocal of the time interval. \[ \text{Frequency} = \frac{1}{0.75} \approx 1.33 \text{ blows per second} \]
02

Understand the Concept of Beats in Hammering

The beat frequency is defined by the condition that both carpenters strike together every 4.6 seconds. The beat frequency is therefore the reciprocal of this time period. \[ \text{Beat Frequency} = \frac{1}{4.6} \approx 0.217 \text{ beats per second} \]
03

Determine the Second Carpenter's Hammering Frequencies

The difference in hammering frequencies between the two carpenters equals the beat frequency. If the second carpenter hammers more rapidly, then: \[ f_2 = f_1 + \text{Beat Frequency} = 1.33 + 0.217 \approx 1.547 \text{ blows per second} \]
04

Calculate the Time Interval for Faster Hammering

If the second carpenter hammers more rapidly, the time interval is the reciprocal of his calculated frequency: \[ \text{Time Interval} = \frac{1}{1.547} \approx 0.65 \text{ seconds} \]
05

Determine the Second Carpenter's Hammering Frequency for Slower Hammering

If the second carpenter hammers less rapidly, then: \[ f_2 = f_1 - \text{Beat Frequency} = 1.33 - 0.217 \approx 1.113 \text{ blows per second} \]
06

Calculate the Time Interval for Slower Hammering

If the second carpenter hammers less rapidly, the time interval is the reciprocal of his calculated frequency: \[ \text{Time Interval} = \frac{1}{1.113} \approx 0.90 \text{ seconds} \]

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Key Concepts

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

Hammering Frequency
When dealing with problems involving hammering, the term "hammering frequency" is exactly what its name suggests: the frequency or rate at which hammer blows are delivered over a period of time. In our example, each carpenter has a distinct hammering frequency. The hammering frequency is calculated as the reciprocal of the time interval between consecutive hammer strokes.
For instance, if a carpenter takes 0.75 seconds to deliver a hammer blow, then his hammering frequency can be calculated as follows:
  • The hammering frequency = \( \frac{1}{0.75} \approx 1.33 \) blows per second.
Understanding this concept helps in solving problems where the regularity of actions needs to be coordinated, like in construction tasks where timing can affect work efficiency and safety.
Time Interval
The time interval in hammering classic problems refers to the duration between each blow from the carpenter. When determining how often the carpenters strike together (besides solving specific problems), understanding this interval is crucial.
For example, let's consider two carpenters who synchronize their strikes every 4.6 seconds, creating a hammering beat frequency.
The first carpenter strikes every 0.75 seconds, established by:
  • Determining his frequency as 1.33 blows per second.
For the second carpenter, whether he hammers faster or slower than his partner affects the time interval between his blows. This time interval is the reciprocal of his hammering frequency. Hence, if his frequency is higher, the interval is shorter than 0.75 seconds, and if slower, the interval is longer. Understanding this concept helps you synchronize tasks that involve rhythmic patterns.
Reciprocal Calculation
Reciprocal calculations are fundamental in determining frequency and time interval relationships, and are heavily employed in our hammering problem. The reciprocal of a number is essentially one divided by that number. In this context, it's used to switch between talking about frequency and time intervals.
As a critical exercise component, calculate the hammering frequency by taking the reciprocal of the time taken to deliver one hammer blow.
For a carpenter who strikes every 0.75 seconds, the reciprocal calculation becomes:
  • The hammering frequency = \( \frac{1}{0.75} \approx 1.33 \) blows per second.
Reciprocals also help compute the time interval for a particular hammering frequency.
For instance, if another carpenter’s frequency is 1.55 blows per second, her time interval for each strike can be calculated as the reciprocal of her frequency:
  • Time interval = \( \frac{1}{1.55} \approx 0.65 \) seconds.
This concept is essential for a mathematical understanding of frequency dynamics in real-world scenarios, such as construction or even musical beats.

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Most popular questions from this chapter

Two ideal gases have the same temperature and the same value for \(\gamma\) (the ratio of the specific heat capacities at constant pressure and constant volume). A molecule of gas A has a mass of \(7.31 \times 10^{-26} \mathrm{kg}\) and a molecule of gas \(\mathrm{B}\) has a mass of \(1.06 \times 10^{-25} \mathrm{kg} .\) When gas \(\mathrm{A}\) (speed of sound \(=259 \mathrm{m} / \mathrm{s}\) ) fills a tube that is open at both ends, the first overtone frequency of the tube is \(386 \mathrm{Hz}\). Gas \(\mathrm{B}\) fills another tube open at both ends, and this tube also has a first overtone frequency of \(386 \mathrm{Hz}\) What is the length of the tube filled with gas B?

A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of \(425 \mathrm{Hz},\) while the piece open only at one end has a fundamental frequency of \(675 \mathrm{Hz}\). What is the fundamental frequency of the original tube?

A pipe open only at one end has a fundamental frequency of 256 Hz. A second pipe, initially identical to the first pipe, is shortened by cutting off a portion of the open end. Now, when both pipes vibrate at their fundamental frequencies, a beat frequency of \(12 \mathrm{Hz}\) is heard. How many centimeters were cut off the end of the second pipe? The speed of sound is \(343 \mathrm{m} / \mathrm{s}\)

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A vertical tube is closed at one end and open to air at the other end. The air pressure is \(1.01 \times 10^{5} \mathrm{Pa}\). The tube has a length of \(0.75 \mathrm{m} .\) Mercury (mass density \(\left.=13600 \mathrm{kg} / \mathrm{m}^{3}\right)\) is poured into it to shorten the effective length for standing waves. What is the absolute pressure at the bottom of the mercury column, when the fundamental frequency of the shortened, air-filled tube is equal to the third harmonic of the original tube?

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