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.