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A tuning fork A, marked 512 Hz, produces...

A tuning fork A, marked 512 Hz, produces 5 beats per second, when sounded with another unmarked tuning fork B. If B is loaded with wax, the number of beats is again 5 per second. What is the frequency of tuning fork B when not loaded? a) 502 Hz b) 507 Hz c) 517 Hz d) 522 Hz

A

502 Hz

B

507 Hz

C

517 Hz

D

522 Hz

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine the frequency of tuning fork B when it is not loaded. Let's break down the solution step by step. ### Step 1: Identify the given data - Frequency of tuning fork A (n_A) = 512 Hz - Number of beats per second (m) = 5 beats/s ### Step 2: Understand the concept of beats When two tuning forks are sounded together, the number of beats per second is equal to the absolute difference between their frequencies. Therefore, we can express the frequency of tuning fork B (n_B) when it is not loaded as: \[ n_B = n_A \pm m \] ### Step 3: Calculate the possible frequencies of tuning fork B Substituting the known values into the equation: 1. When n_B is higher than n_A: \[ n_B = 512 + 5 = 517 \text{ Hz} \] 2. When n_B is lower than n_A: \[ n_B = 512 - 5 = 507 \text{ Hz} \] Thus, the possible frequencies for tuning fork B when it is not loaded are 517 Hz and 507 Hz. ### Step 4: Analyze the effect of loading fork B with wax When tuning fork B is loaded with wax, the frequency decreases. The problem states that the number of beats remains 5 beats/s when fork B is loaded. This means that the frequency of the loaded fork (n_B') is still either: - \( n_B' = 512 + 5 = 517 \text{ Hz} \) (if n_B was 512 Hz) - \( n_B' = 512 - 5 = 507 \text{ Hz} \) (if n_B was 517 Hz) Since the number of beats is the same (5 beats/s), we can conclude that the frequency of B when loaded must be 507 Hz (which is lower than 512 Hz). ### Step 5: Determine the frequency of tuning fork B when not loaded Since the loaded frequency of B is 507 Hz, and this corresponds to the scenario where the original frequency of B was 512 Hz (the only way to get 5 beats with A at 512 Hz), we conclude that: \[ n_B = 517 \text{ Hz} \] ### Final Answer Thus, the frequency of tuning fork B when not loaded is: **517 Hz (Option c)** ---
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