When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard.

Q: Why does adding tape decrease the frequency?

Adding tape increases the mass of the tuning fork's prongs without changing their stiffness. A higher mass means more inertia, causing the prongs to oscillate more slowly, which lowers the frequency.

Q: What would happen if we filed the tuning fork instead of taping it?

Filing the prongs removes material, which decreases their mass. With less mass and less inertia, the prongs would vibrate faster, resulting in an increased frequency.

Q: Could the 204 Hz fork drop all the way to 194 Hz to give 6 beats?

Mathematically, yes. But physically, a piece of tape usually causes a very small drop in frequency (like 1 or 2 Hz). A massive drop of 10 Hz is highly unrealistic for just adding tape.

Q: Why are there two possible frequencies initially?

Because the beat frequency only tells us the *difference* between the two frequencies, not which one is higher. An absolute value equation $|f_1 - f_2| = 4$ always yields two solutions.

Q: Does the frequency of fork 1 change during the experiment?

No, fork 1 is left completely untouched. It acts as our constant reference point of 200 Hz throughout the entire process.

Question

When two tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats per second are heard. Now, some tape is attached on the prong of the fork 2. When the tuning forks are sounded again, 6 beats per second are heard.
If the frequency of fork 1 is 200 Hz, then what will be the original frequency of fork 2?

Options:

(A) 200 Hz

(B) 202 Hz

(C) 196 Hz

(D) 204 Hz

Sigma Percentile · Accepted Answer

The phenomenon of beats is one of the most fascinating consequences of wave interference. When two sound waves of slightly different frequencies are played together, they periodically construct and destruct, creating a pulsing sound.

In this problem, we are acting as acoustic detectives. We have two tuning forks, and we need to deduce the original frequency of the second one using only the beat frequencies.

### The Beat Phenomenon

Let's start by analyzing the initial setup. We are given that fork 1 has a known frequency.

$$f_1 = 200 	ext{ Hz}$$

When sounded together with fork 2, we hear 4 beats per second. The beat frequency is simply the absolute difference between the two frequencies.

$$f_{beat} = |f_1 - f_2| = 4 	ext{ Hz}$$

### The Two Possibilities

Because of the absolute value in our equation, there are two mathematical possibilities for the frequency of fork 2. It could either be 4 Hz higher than fork 1, or 4 Hz lower.

$$f_2 = 200 + 4 = 204 	ext{ Hz}$$

$$f_2 = 200 - 4 = 196 	ext{ Hz}$$

At this stage, both 204 Hz and 196 Hz are equally valid candidates. To find the true original frequency, we need to introduce a physical change and observe the outcome.

### The Tape Effect

The problem states that we attach some tape to the prong of fork 2. What does this do physically?

Adding tape increases the mass of the tuning fork's prongs. Just like a heavier person on a swing oscillates more slowly, a heavier tuning fork vibrates at a lower frequency. Therefore, we know for certain that the new frequency of fork 2, let's call it $f_2'$, must be less than its original frequency.

$$f_2' < f_2$$

After adding the tape, the new beat frequency increases to 6 beats per second. Let's test our two candidates against this new evidence.

### Solving the Mystery

**Case 1: What if the original frequency was 204 Hz?**

If $f_2 = 204 	ext{ Hz}$, adding tape will decrease its frequency. It will drop to 203, 202, or 201 Hz. Notice that as it decreases, it gets closer to the 200 Hz of fork 1.

This means the gap between the two frequencies shrinks, and the beat frequency would decrease. However, the problem states the beats increased to 6. Therefore, 204 Hz is physically impossible.

**Case 2: What if the original frequency was 196 Hz?**

If $f_2 = 196 	ext{ Hz}$, adding tape will decrease its frequency further, perhaps to 195 or 194 Hz. As it decreases, it moves further away from the 200 Hz of fork 1.

The gap widens! If the frequency drops specifically to 194 Hz, the difference between 200 Hz and 194 Hz is exactly 6 Hz.

$$|200 - 194| = 6 	ext{ Hz}$$

This perfectly matches the condition given in the problem. Thus, we have solved the mystery. The original frequency of the second tuning fork must have been exactly 196 Hz.

Select Answer:

Visualized Solution

$f_{1}$ Anchor

Beat Frequency

Possible Values

Effect of Tape

Testing $204 Hz$

Testing $196 Hz$

Final Answer

The Sigma Insight: Interference of Waves

The Beat Phenomenon

The Two Possibilities

The Tape Effect

Solving the Mystery

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