A tuning fork A, marked 512 Hz, produces 5 beats per second, where sounded with another unmarked tuning fork B

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Q. A tuning fork A, marked 512 Hz, produces 5 beats per second, where 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 the tuning fork B when not loaded?

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Concept understanding — Beats Frequency Analysis

Beats Frequency Analysis

The Intuition: When Two Tones "Wobble"

Imagine you're tuning a guitar. You pluck the string you're tuning, and at the same time, you play a reference note from a tuning fork. If the two notes are exactly the same pitch, you hear a single steady tone. But if they are slightly different — say one is 440 Hz and the other is 442 Hz — you don't hear two separate notes. Instead, you hear a single tone that wobbles in loudness: it gets louder, then softer, then louder again, in a slow, rhythmic pulse.

That pulse is called a beat. The phenomenon is beats.

Why does this happen? Because the two sound waves are constantly going in and out of sync. When their crests align, they add up to a louder sound (constructive interference). When a crest meets a trough, they cancel partially (destructive interference). The result is a wave whose amplitude rises and falls at a rate equal to the difference between the two original frequencies.

Note

You don't hear the individual frequencies when they are very close. Your ear perceives the average frequency (around 441 Hz in the example), but the loudness fluctuates at the beat frequency.

The Precise Statement

Let two sound waves of slightly different frequencies $f_1$ and $f_2$ (with $f_1 > f_2$ ) travel through the same medium. Their displacements at a point can be written as:

y_1 = A \sin(2\pi f_1 t)

y_2 = A \sin(2\pi f_2 t)

By the principle of superposition, the resultant displacement is:

y = y_1 + y_2 = A \left[ \sin(2\pi f_1 t) + \sin(2\pi f_2 t) \right]

Using the trigonometric identity $\sin P + \sin Q = 2 \sin\left(\frac{P+Q}{2}\right) \cos\left(\frac{P-Q}{2}\right)$ , we get:

y = 2A \cos\left( 2\pi \frac{f_1 - f_2}{2} t \right) \sin\left( 2\pi \frac{f_1 + f_2}{2} t \right)

This is the key equation. It describes a wave with two parts:

The carrier wave: $\sin\left( 2\pi \frac{f_1 + f_2}{2} t \right)$ — this oscillates at the average frequency $f_{\text{avg}} = \frac{f_1 + f_2}{2}$ . This is the pitch you actually hear.
The envelope: $2A \cos\left( 2\pi \frac{f_1 - f_2}{2} t \right)$ — this modulates the amplitude of the carrier. The envelope oscillates at half the difference frequency.

f_{\text{beat}} = |f_1 - f_2|

Why $|f_1 - f_2|$ and not half of it? Because the cosine term goes through a full cycle (from maximum to minimum and back to maximum) when its argument changes by $2\pi$ . That happens when $\frac{f_1 - f_2}{2} t = 1$ , i.e., $t = \frac{2}{f_1 - f_2}$ . So the time period of the envelope is $T_{\text{envelope}} = \frac{2}{|f_1 - f_2|}$ . But the loudness (intensity) goes through two maxima per envelope cycle — one at each positive peak of the cosine and one at each negative peak (since squaring the amplitude gives intensity). So the period of the beat (the time between successive loudness maxima) is half of that: $T_{\text{beat}} = \frac{1}{|f_1 - f_2|}$ . Hence, the beat frequency is $f_{\text{beat}} = \frac{1}{T_{\text{beat}}} = |f_1 - f_2|$ .

Watch out

A common mistake is to think the beat frequency is $\frac{|f_1 - f_2|}{2}$ . That is the frequency of the envelope oscillation, not the beat. The ear detects two loudness peaks per envelope cycle, so the beat frequency is double the envelope frequency.

Key Conditions for Beats

Small difference: The two frequencies must be close (typically less than about 10–15 Hz apart). If the difference is too large, the ear perceives two separate tones instead of beats.
Comparable amplitudes: The amplitudes should be roughly equal for maximum contrast in loudness. If one is much louder, beats are still present but less noticeable.
Same medium: The waves must overlap in the same region of space.

Why This Matters for Exams …

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When two sound sources of slightly different frequencies are sounded together, they produce beats. The beat frequency is the absolute difference between their individual frequencies. Loading a tuning fork with wax decreases its frequency. By analyzing the initial and final beat frequencies, we deduce that the frequency of tuning fork B when not loaded is $\boxed{517 \text{ Hz}}$ .

When two sound waves of slightly different frequencies interfere, they produce a phenomenon called beats. This is perceived as a periodic variation in the loudness of the sound. The number of beats heard per second is known as the beat frequency.

The core idea behind solving this problem is to understand:

How beat frequency is calculated.
How loading a tuning fork with wax affects its frequency.

Let's break down the concepts and then apply them.

Concept: Beats and Frequency

When two sound waves with frequencies $f_1$ and $f_2$ are superposed, the beat frequency ( $f_{beats}$ ) is given by:

f_{beats} = |f_1 - f_2|

This means that if you hear $N$ beats per second, the frequency of the unknown source can be either $N$ Hz higher or $N$ Hz lower than the known source.

Concept: Effect of Loading a Tuning Fork

A tuning fork vibrates at its natural frequency. When a small amount of wax is added to the prongs of a tuning fork, it increases the effective mass of the vibrating system. An increase in mass leads to an increase in inertia, which in turn decreases the natural frequency of vibration.

Important

Loading a tuning fork with wax always decreases its frequency.

Now, let's apply these concepts to the problem.

Step-by-Step Solution

Determine the initial possible frequencies of tuning fork B.
- We are given the frequency of tuning fork A, $f_A = 512$ Hz.
- The number of beats produced when A and B are sounded together is 5 beats per second. So, $f_{beats} = 5$ Hz.
- Using the beat frequency formula, $f_{beats} = |f_A - f_B|$ , we have: $5 = |512 - f_B|$
- This gives us two possibilities for the frequency of tuning fork B ( $f_B$ $f_{B}$ ):
  - $512 - f_B = 5 \implies f_B = 512 - 5 = 507$ Hz
  - $512 - f_B = -5 \implies f_B = 512 + 5 = 517$ Hz
- So, initially, $f_B$ is either 507 Hz or 517 Hz.
Analyze the effect of loading tuning fork B with wax.
- When tuning fork B is loaded with wax, its frequency decreases. Let the new frequency of B be $f_B'$ .
- Therefore, $f_B' < f_B$ .
Determine the new possible frequencies of tuning fork B after loading.
- After loading B with wax, the number of beats is again 5 per second.
- So, the new beat frequency $f_{beats}' = 5$ Hz.
- Using the beat frequency formula with the new frequency $f_B'$ : $5 = |f_A - f_B'|$ $5 = |512 - f_B'|$
- This again gives two possibilities for $f_B'$ $f_{B}^{'}$ :
  - $512 - f_B' = 5 \implies f_B' = 512 - 5 = 507$ Hz
  - $512 - f_B' = -5 \implies f_B' = 512 + 5 = 517$ Hz
- So, after loading, $f_B'$ is either 507 Hz or 517 Hz.
Reconcile the initial and final frequencies to find the correct $f_B$ .

We need to find an initial frequency $f_B$ from Step 1 such that when it decreases (as per Step 2), it matches one of the possible $f_B'$ values from Step 3.
- Case 1: Assume initial $f_B = 507$ Hz.
  - If $f_B = 507$ Hz, then after loading with wax, $f_B'$ must be less than 507 Hz.
  - However, the possible values for $f_B'$ (from Step 3) are 507 Hz or 517 Hz. Neither of these is strictly less than 507 Hz. …

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Q.A tuning fork A, marked 512 Hz, produces 5 beats per second, where 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 the tuning fork B when not loaded?

Concept: Beats and Frequency

Concept: Effect of Loading a Tuning Fork

Step-by-Step Solution

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