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Chapter 16: Problem 97

Two running forks \(P\) and \(Q\) when set vibrating, given 4 beats \(\mathrm{s}^{-1}\). If a prong of the fork \(P\) is filed, the beats are reduced to \(2 / \mathrm{s}\). What is frequency of \(P\), if that of \(Q\) is \(250 \mathrm{~Hz} ?\) (a) \(246 \mathrm{~Hz}\) (b) \(250 \mathrm{~Hz}\) (c) \(254 \mathrm{~Hz}\) (d) \(252 \mathrm{~Hz}\)

Short Answer

Expert verified
The frequency of fork P is 254 Hz (option c).

Step by step solution

01

Understand the Concept of Beats

Beats occur when two sound waves of similar frequencies interfere with each other. The beat frequency, which is the number of beats per second, is the absolute difference between the two frequencies of the waves. In this case, the running forks P and Q produce 4 beats per second.
02

Write the Equation for Beat Frequency

Let the frequency of fork P be \(f_P\). Given that the frequency of fork Q is \(f_Q = 250 \, \text{Hz}\), the equation for the beat frequency is \( |f_P - f_Q| = 4 \, \text{Hz}\). This simplifies to two possibilities: \( f_P - 250 = 4 \) or \( 250 - f_P = 4 \).
03

Solve for Initial Frequency of P

From Step 2, two scenarios arise: (1) \( f_P = 250 + 4 = 254 \, \text{Hz}\), and (2) \( f_P = 250 - 4 = 246 \, \text{Hz}\). Initially, \(f_P\) could be either 254 Hz or 246 Hz.
04

Analyze the Scenario when Fork P is Filed

Filing fork P reduces its frequency. The new beat frequency is 2 beats per second when fork P is filed. The equation becomes \( |f'_P - 250| = 2 \, \text{Hz}\), where \( f'_P \) is the reduced frequency of fork P.
05

Determine Suitable Frequency for Filed Fork P

Initially, consider \( f_P = 246 \, \text{Hz}\). If fork P is filed, its frequency must decrease further, so it cannot satisfy \( |f'_P - 250| = 2 \). Now, consider \( f_P = 254 \, \text{Hz}\). With filing, assume \( f'_P = 252 \, \text{Hz}\), which satisfies \( 254 - 250 = 4 \) initially and \( 252 - 250 = 2 \) when filed.
06

Confirm Consistency

Verify against the original conditions: The unfiled fork, with \( f_P = 254 \, \text{Hz}\), creates 4 beats per second with Q (250 Hz). When filed to 252 Hz, it creates 2 beats per second with Q. This confirms consistency with the problem's conditions.

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Key Concepts

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

Interference of Sound Waves
When we talk about the interference of sound waves, we're mainly discussing how two sound waves interact with each other. This interaction can lead to different phenomena, one of which is the creation of beats. This happens when two sound waves with frequencies that are close, but not exactly the same, overlap.

The interference causes the amplitude of the resultant wave to oscillate over time. This oscillation is perceived by our ears as a variation in volume, or pulsating sound, which are called "beats." The beat frequency is determined by the absolute difference between the frequencies of the two initial waves.

For example, in the problem mentioned, the waves from the tuning forks P and Q interfere, and the beats are heard at a frequency of 4 beats per second. This simply means that the difference between the frequencies of these two sound waves was 4 Hz initially.

  • Sound waves interfere when they have slightly different frequencies.
  • Beat frequency is the difference in frequencies of interfering sound waves.
  • Beats are perceived as fluctuations in volume.
Frequency Change Due to Filing
In the scenario provided, one fork is filed, affecting its frequency. Filing a tuning fork reduces its length, which in turn reduces its frequency. Imagine filing down one prong of a tuning fork. The reduction in its physical dimensions means that it will vibrate at a slower rate.

When such physical alterations are applied, the frequency of sound emitted by the fork changes. This principle is critical in the given exercise because the change in frequency after filing is what influences the beat frequency heard.

After filing fork P, the beats reduce from 4 per second to 2 per second. This implies that the change in the frequency of fork P corrected the difference in frequency with fork Q, leading to a new beat frequency. This reduction is key to determining the original frequency of fork P.

  • Filing a tuning fork decreases its frequency.
  • The resulting frequency change alters the beat frequency.
  • Certain physical changes directly impact sound frequency.
Sound Wave Frequency Calculation
Calculating sound wave frequency involves understanding the relationships between the initial and changed frequencies and the resultant beat frequency. In this specific problem, understanding these relationships is crucial to solve for the frequency of fork P.

Initially, we set up an equation for the original beat frequency: \[ |f_P - 250| = 4 \]where \(f_P\) is the frequency of fork P, and the frequency of fork Q is known to be 250 Hz.

From this equation, two possible values of \(f_P\) arise: 254 Hz or 246 Hz. Filing lowers \(f_P\), which helps eliminate the possibility of \(f_P = 246 \) Hz, since the further reduction would not result in 2 beats per second. Checking the remaining option, if \(f_P = 254 \) Hz, then after filing, the fork鈥檚 frequency could realistically become 252 Hz, resulting in:\[ |f'_P - 250| = 2 \]confirming its suitability for the conditions given.

  • Original beat frequency calculation: the absolute frequency difference.
  • Use given conditions to deduce correct frequency values.
  • Verify results by recalculating beat frequency after changes.

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Most popular questions from this chapter

The tension in a wire is decreased by \(19 \%\). The percentage decrease in frequency will be (a) \(19 \%\) (b) \(10 \%\) (c) \(0.19 \%\) (d) None of these

In a resonance tube, using a tuning fork of frequency \(325 \mathrm{~Hz}\), two successive resonance length are observed as \(25.4 \mathrm{~cm}\) and \(77.4 \mathrm{~cm}\) respectively. The velocity of sound in air is (a) \(338 \mathrm{~ms}^{-1}\) (b) \(328 \mathrm{~ms}^{-1}\) (c) \(330 \mathrm{~ms}^{-1}\) (d) \(320 \mathrm{~ms}^{-1}\)

The amplitude of wave disturbance propagating in positive direction of \(X\)-axis is given by \(y=\frac{1}{1+x^{2}}\) at \(t=0\) and by \(y=\frac{1}{1+(x-1)^{2}}\) at \(t=2 \mathrm{~s}\) where \(x\) and \(y\) are in metres. The shape of the wave disturbance does not change during propagation. The velocity of the wave is. (a) \(0.5 \mathrm{~ms}^{-1}\) (b) \(2.0 \mathrm{~ms}^{-1}\) (c) \(1.0 \mathrm{~ms}^{-1}\) (d) \(4.0 \mathrm{~ms}^{-1}\)

A fork of unknown frequency gives four beats \(\mathrm{s}^{-1}\) when sounded with another of frequency 256 . The fork is now loaded with a piece of wax and again four beats \(\mathrm{s}^{-1}\) are heard. Then the frequency of the unknown fork is (a) \(256 \mathrm{~Hz}\) (b) \(252 \mathrm{~Hz}\) (c) \(264 \mathrm{~Hz}\) (d) \(260 \mathrm{~Hz}\)

An open pipe is suddenly closed at one end with the result that the frequency of third harmonic of the closed pipe is found to be higher at \(100 \mathrm{~Hz}\). The fundamental frequency of the open pipe is (a) \(200 \mathrm{~Hz}\) (b) \(480 \mathrm{~Hz}\) (c) \(240 \mathrm{~Hz}\) (d) \(300 \mathrm{~Hz}\)
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