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Chapter 12: Problem 51

A number of tuning forks are arranged in the order of increasing frequency and any two successive tuning forks produce 4 beats per second, when sounded together. If the last tuning fork has a frequency octave higher than that of the first tuning fork and the frequency of the first tuning fork is \(256 \mathrm{~Hz}\), then the number of tuning forks is (a) 63 (b) 64 (c) 65 (d) 66

Short Answer

Expert verified
The number of tuning forks is 65.

Step by step solution

01

Understanding the Problem

We are given a sequence of tuning forks with increasing frequencies. The first fork has a frequency of 256 Hz. Each subsequent fork produces 4 beats per second with the previous one, meaning the frequency difference between successive forks is 4 Hz. The last fork's frequency is an octave higher than the first.
02

Defining an Octave

An octave is defined as a doubling of frequency. If the first tuning fork has a frequency of 256 Hz, then an octave higher means the last tuning fork has a frequency of 512 Hz (since 256 Hz 脳 2 = 512 Hz).
03

Calculating the Number of Intervals

By the problem, each successive tuning fork differs by 4 Hz. Starting from 256 Hz and increasing to 512 Hz, we need to find how many such 4 Hz jumps can occur. The total frequency increase is 512 - 256 = 256 Hz.
04

Determining the Number of Forks

Each increase is 4 Hz, so there are 256 Hz / 4 Hz = 64 intervals. Since these are intervals between the forks, there will be 64 + 1 = 65 tuning forks in total (starting with the first fork and including the last one).
05

Answer Selection

Given the calculation, the correct number of tuning forks is 65, which corresponds to answer option (c).

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Key Concepts

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

Beats per second
In the world of sound and music, understanding the concept of "beats per second" can be quite enlightening. When two sound waves of slightly differing frequencies interfere, they create a pattern of beats. These beats are moments of constructive interference where the sound is perceived as being louder. The frequency of these beats is known as the "beat frequency."
  • When you have two frequencies, say, \( f_1 \) and \( f_2 \), the beat frequency \( f_{b} \) is given by \( |f_1 - f_2| \).
  • In the context of tuning forks, if two successive forks produce 4 beats per second when sounded together, then the difference in their frequencies is 4 Hz.
This concept is pivotal in tuning musical instruments and in our exercise as it helps determine the frequency interval between tuning forks.
Octave definition
An octave is a fundamental concept in music theory and physics. It describes the interval between one musical pitch and another with double its frequency.
  • If a tuning fork has a frequency of 256 Hz, its octave would be 512 Hz, doubling its original frequency.
  • This doubling of frequency means that sounds an octave apart are perceived as being very similar in tone.
Understanding octaves is crucial in the exercise since the final tuning fork is stated to have a frequency one octave higher, framing our problem in mathematical terms of frequency increase.
Frequency intervals
Frequency intervals are the differences between successive sound frequencies. These intervals help us quantify the separation between different pitches. In this particular problem, the tuning forks are arranged so each successive fork has a frequency interval of 4 Hz compared to the previous one.
  • The interval formula used here is simple: if each fork has a difference of 4 Hz from the previous one, it aligns with our understanding of beat frequency.
  • This creates a consistent increase in frequency by 4 Hz, which is the key to figuring out the total number of intervals needed to span from 256 Hz to 512 Hz.
Calculating these intervals leads us to conclude the necessary count of tuning forks in the sequence.
NEET physics problem
The NEET exam often challenges students with physics problems that combine theoretical concepts with practical applications. This exercise is a great example of such a problem, blending musical physics concepts, like frequency and sound waves, with mathematical calculations.
  • To succeed in such problems, a firm understanding of concepts like frequency intervals, beat frequency, and the definition of an octave is crucial.
  • Engaging with these problems develops critical thinking and the ability to apply theoretical knowledge to solve practical issues, which is a key skill for anyone preparing for competitive exams like NEET.
By breaking down problems as shown here, students can better prepare for the type of multifaceted questions they will encounter on their exams.

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Most popular questions from this chapter

If a note \(x\) of unknown frequency produces 8 beats/ sec, with a source of \(250 \mathrm{~Hz}\) and 12 beats/sec with a source of \(270 \mathrm{~Hz}\), the frequency of unknown source will be (a) \(258 \mathrm{~Hz}\) (b) \(242 \mathrm{~Hz}\) (c) \(262 \mathrm{~Hz}\) (d) \(282 \mathrm{~Hz}\)

Apparatus used to find out velocity of sound in gas is (a) Melde's apparatus (b) Kundt's tube (c) Quincke's tube (d) none of these

When beats are produced by two progressive waves of nearly the same frequency, which one of the following is correct? (a) The particles vibrate simple harmonically, with the frequency equal to the difference in the component frequencies (b) The amplitude of vibrations at any point changes simple harmonically with a frequency equal to difference in the frequencies of the two waves (c) The frequency of the beats depends on the position, where the observer is (d) The frequency of the beat changes as the time progresses (e) The particle's vibration frequency and amplitude frequency are equal

Three sound waves of equal amplitude have frequencies \((v+1), v,(v+1)\). They superpose to give beats. The number of beats produced per second will be (a) 4 (b) 3 (c) 2 (d) 1

A blast gives a sound of intensity \(0.8 \mathrm{~W} / \mathrm{m}^{2}\) and frequency \(1 \mathrm{kHz}\). If the density of air is \(1.3 \mathrm{~kg} / \mathrm{m}^{3}\) and speed of sound in air is \(330 \mathrm{~m} / \mathrm{s}\), the amplitude of the sound wave is approximately (a) \(5 \times 10^{-6} \mathrm{~m}\) (b) \(9.7 \times 10^{-6} \mathrm{~m}\) (c) \(15 \times 10^{-6} \mathrm{~m}\) (d) \(20 \times 10^{6} \mathrm{~m}\)
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