Animated Solution for Physics - Waves: A tuning fork A of unknown frequency produces 5 beats/s with a fork of known frequency 340 Hz. When fork A is filled, the beat frequency decreases to 2 beats/s. What is the frequency of fork A?

Imagine you are a detective, and your job is to find the exact frequency of a mysterious tuning fork. You are given a few clues: a known tuning fork, an initial beat frequency, and a physical modification made to the unknown fork. Let's break down this classic physics mystery step by step!

Our Anchor Point

We start by visualizing the frequencies on a number line. We have a known tuning fork, let's call it Fork B, with a frequency of exactly $340\text{ Hz}$. This is our anchor point.

The problem tells us that our unknown Fork A produces $5\text{ beats/s}$ with Fork B. We know that the beat frequency is simply the absolute difference between the frequencies of the two tuning forks:

Since the beat frequency is $5\text{ Hz}$, our unknown Fork A could either be $5\text{ Hz}$ above or $5\text{ Hz}$ below Fork B.

This gives us two possible suspects for the frequency of Fork A: it is either sitting at $345\text{ Hz}$ or $335\text{ Hz}$. To find out which one is the true frequency, we need to look at the next clue.

Filing the Fork

The problem states that Fork A is "filled". In the context of physics problems, this is almost always a typographical error for filed. When you file a tuning fork, you scrape off some of the metal from its prongs.

Think of a tuning fork like a mass on a spring. The frequency of oscillation is inversely proportional to the square root of the mass ($f\propto \frac{1}{\sqrt{m}}$). By filing the fork, we decrease its mass, which causes it to vibrate faster. Therefore, filing a tuning fork always increases its frequency. On our number line, this means the frequency of Fork A will shift to the right.

Let's test our first suspect, $345\text{ Hz}$. If we file it, its frequency increases, moving further to the right (e.g., to $347\text{ Hz}$ or $348\text{ Hz}$). The gap between it and our anchor at $340\text{ Hz}$ will now be strictly greater than $5\text{ Hz}$. But the question says the new beat frequency decreases to $2\text{ Hz}$! So, $345\text{ Hz}$ is definitely not our answer.

Now let's look at the second suspect, $335\text{ Hz}$. If we file this one, its frequency increases, moving it to the right and closer to $340\text{ Hz}$. The gap shrinks! If the new frequency reaches $338\text{ Hz}$, the new beat frequency would be exactly:

This perfectly matches our given condition.

By eliminating the impossible, we can confidently conclude that the original frequency of tuning fork A must have been $335\text{ Hz}$.

As a mental exercise, think about what would happen if the problem said we added wax to the tuning fork instead of filing it. Adding wax increases mass and decreases frequency, which would completely flip our number line analysis!