Question

A tuning fork arrangement produces 4 beats/second with one fork of frequency \(288 \mathrm{~Hz}\). A little wax is applied on the unknown fork and it then produces 2 beats/s. The frequency of the unknown fork is \(\ldots \ldots \ldots . \mathrm{Hz}\). (A) 286 (B) 292 (C) 294 (D) 288

Short answer

The frequency of the unknown fork after applying the wax is \(286 \thinspace Hz\). Therefore, the correct answer is (A) 286 Hz.

Step-by-step solution

Write the initial beat frequency formula

The beat frequency before applying the wax is 4 beats/s. The formula to compute the beat frequency is \(beat \thinspace frequency = |f_1 - f_2|\), where \(f_1\) and \(f_2\) are the frequencies of the two forks. In this case, we have \(f_1 = 288 \thinspace Hz\) and \(f_2\) is the frequency of the unknown fork. The initial beat frequency can be represented as: \( |288 - f_2| = 4 \) #Step 2: Find the beat frequency after applying the wax#

Write the beat frequency formula after applying the wax

The beat frequency after applying the wax is 2 beats/s. When the wax is applied, the frequency of the unknown fork increases. Now, we'll represent the new beat frequency as: \( |288 - (f_2 + w)| = 2 \) Where \(w\) represents the increase in frequency due to the wax. #Step 3: Solve the system of equations#

Solve the given system of equations

We now have a system of equations: 1. \( |288 - f_2| = 4 \) 2. \( |288 - (f_2 + w)| = 2 \) We will now consider two cases for each equation: For Eq.1: (a) \( 288 - f_2 = 4 \) (b) \( f_2 - 288 = 4 \) For Eq.2: (c) \( 288 - (f_2 + w) = 2 \) (d) \( (f_2 + w) - 288 = 2 \) We need to find one pair of equations that makes sense with the given information. In this problem, we know that adding wax increases the frequency of the unknown fork, so we can rule out Eq.(c) because it would indicate that adding wax decreases the frequency. Hence, we focus on Eq.(b) and Eq.(d). From Eq.(b): \( f_2 = 288 + 4 \) \( f_2 = 292 \thinspace Hz \) From Eq.(d): \( f_2 + w = 288 + 2 \) \( 292 + w = 290 \) \( w = -2 \) This is not possible because the frequency should increase after adding wax. So, the only feasible solution is Eq.(a) and Eq.(c). From Eq.(a): \( 288 - f_2 = 4 \) \( f_2 = 288 - 4 \) \( f_2 = 284 \thinspace Hz \) From Eq.(c): \( 288 - (f_2 + w) = 2 \) \( 288 - (284 + w) = 2 \) \( 4 - w = 2 \) \( w = 2 \) #Step 4: Find the unknown fork's frequency#

Calculate the final frequency of the unknown fork

Now that we have the frequency of the unknown fork before wax addition (284 Hz) and the increase in frequency due to the wax (2 Hz), we can find the final frequency of the unknown fork: \( f_2\_final = f_2 + w \) \( f_2\_final = 284 + 2 \) \( f_2\_final = 286 \thinspace Hz \) This means that the frequency of the unknown fork after applying the wax is 286 Hz. Therefore, the correct answer is (A) 286 Hz.

Key concepts

beat frequency

The concept of beat frequency arises when two sound waves of slightly different frequencies interfere with each other. As these two waves overlap, they create a new wave pattern characterized by alternating constructive and destructive interference. This results in the perception of beats. Essentially, beats are the rhythmic pulsing heard when two tones of a close, but not identical frequency, are played together.
The beat frequency is the difference between the frequencies of these two waves. It can be calculated using the formula:

• \(\text{beat frequency} = |f_1 - f_2|\)

Here, \(f_1\) and \(f_2\) represent the frequencies of the two sound sources, like tuning forks.
In the example from the exercise, a tuning fork and an unknown fork produce 4 beats per second, meaning that the frequency difference between them is 4 Hz. Understanding how beats work is crucial when trying to determine unknown frequencies, as it provides a way to measure the frequency difference audibly.

sound waves

Sound waves are vibrations that travel through the air, or any other medium, as an audible mechanical wave. These waves are created by vibrating objects, such as tuning forks, which cause the surrounding medium (usually air) to vibrate at the same frequency. The movement of these waves is what we perceive as sound.
Sound waves share a few important properties:

• Frequency: This determines the pitch of the sound; higher frequencies sound higher-pitched, and lower frequencies sound lower-pitched.
• Amplitude: This relates to the loudness of the sound; larger amplitudes mean louder sounds.
• Wavelength: This is the distance between two consecutive peaks of the wave.

Since the speed of sound is generally constant in a given medium, the frequency and wavelength are inversely related.
When tuning forks oscillate, they disturb the air molecules around them, producing sound waves with specific frequencies. By listening to these frequencies and interactions like beats, we can infer valuable information about sound characteristics and their sources.

resonance

Resonance occurs when a system vibrates at its natural frequency due to an external stimulus, such as sound waves with matching frequencies. When an object is exposed to sound waves that match its resonant frequency, it will absorb energy efficiently, causing amplification and a stronger vibration. This concept can be experienced when loudly playing a note near a glass causes it to vibrate more intensely.
In the context of tuning forks, resonance plays a critical role. When a tuning fork is struck, it vibrates at a specific natural frequency. If another fork of a different frequency is brought nearby, the forks can either dampen or amplify each other's vibrations depending on the proximity of their natural frequencies.
By applying wax to one of the tuning forks in the exercise, the resonance conditions temporarily change — because the frequency of the wax-applied fork shifts slightly, affecting how it interacts with the fixed frequency fork. Understanding resonance helps us predict the resulting frequency shifts and their subsequent beat frequency changes.

frequency calculation

Calculating frequencies, especially in the presence of interacting sound waves, requires a careful approach. In the provided exercise, we are given a tuning fork with a known frequency and an unknown fork that has undergone changes due to wax application. To solve frequency-related problems, it's essential to systematically break down the conditions given in the problem.
Initially, the unknown fork and a 288 Hz fork produced 4 beats per second, leading us to use the beat frequency formula:

• \(|288 - f_2| = 4\)

This provides two possible scenarios \(f_2 = 292\) Hz or \(284\) Hz.
Waxes often change the frequency of forks, so when the beat frequency shifts to 2 beats per second with the wax, it indicates the frequency shift. We then use:

• \(|288 - (f_2 + w)| = 2\)

Solving these equations allows us to assess potential frequency changes due to wax. As detailed solving showed in the exercise, the resolved frequency after all calculations was 286 Hz after adjusting for the wax effect.
Through careful application of beat frequency and resonance principles, along with precise frequency calculations, we can determine the new frequencies of sound-producing objects, helping understand their acoustic behaviors.