Animated Solution for Physics - Waves: A tuning fork arrangement (pair) produces 4 beats per second with one fork of frequency 288 cps. A little wax is placed on the unknown fork and then it produces 2 beats per second. The frequency of the unknown fork is

Imagine you are standing in a room where two musicians are playing slightly different notes on their instruments. Instead of hearing a smooth, continuous sound, you hear a distinct "wobble" or a rhythmic pulsing in the volume. This fascinating acoustic phenomenon is known as beats.

Beats occur due to the principle of superposition. When two sound waves of slightly different frequencies interfere with each other, they periodically fall in and out of phase. When they are in phase, they construct a louder sound (constructive interference). When they are out of phase, they cancel each other out, creating a softer sound (destructive interference).

The number of these loud-soft cycles you hear per second is called the beat frequency. Mathematically, it is beautifully simple. The beat frequency is exactly equal to the absolute difference between the two source frequencies:

In our problem, we are given a known tuning fork with a frequency of $f_1=288\text{ Hz}$ (or cps, which stands for cycles per second). When sounded together with an unknown tuning fork of frequency $f_2$, they produce $4$ beats per second.

Using our beat frequency formula, we can set up the following equation:

Because of the absolute value, this equation branches into two distinct possibilities. The unknown frequency could be $4\text{ Hz}$ higher than the known frequency, or it could be $4\text{ Hz}$ lower.

Possibility 1: $f_2=288+4=292\text{ Hz}$ Possibility 2: $f_2=288-4=284\text{ Hz}$

At this stage, we are at a crossroads. Both $292\text{ Hz}$ and $284\text{ Hz}$ are perfectly valid candidates. We need more physical information to break this tie.

This is where the problem introduces a physical manipulation: placing a little wax on the prongs of the unknown tuning fork. To understand what this does, we must look at the mechanics of a tuning fork.

A tuning fork behaves much like a simple harmonic oscillator, similar to a mass on a spring. Its natural frequency of vibration depends on two main factors: the stiffness of the metal (analogous to the spring constant, $k$) and the mass of the prongs ($m$). The relationship is given by:

When we stick wax onto the prongs, we are effectively increasing the mass ($m$) of the oscillating system without changing its stiffness. According to the proportionality above, an increase in mass in the denominator will cause the overall frequency ($f$) to decrease.

Therefore, whatever the original frequency $f_2$ was, the new frequency $f_2^{\prime }$ after waxing must be strictly less than $f_2$:

Eliminating the Impossible

We are told that after the wax is applied, the new beat frequency drops to $2$ beats per second. Let's test our two initial possibilities against this new constraint.

Testing Possibility 2 ($284\text{ Hz}$): Suppose the original frequency was $284\text{ Hz}$. Adding wax will decrease this frequency further. It might drop to $283\text{ Hz}$, $282\text{ Hz}$, or even lower.

Let's calculate the new beat frequency with the known $288\text{ Hz}$ fork. If the new frequency is, say, $283\text{ Hz}$, the beats would be $|288-283|=5\text{ Hz}$. The gap between the two frequencies is widening! The beat frequency would increase to $5$, $6$, or more.

However, the problem explicitly states the new beat frequency is $2\text{ Hz}$. Since $284\text{ Hz}$ leads to an increase in beats, this possibility is physically impossible and must be rejected.

Testing Possibility 1 ($292\text{ Hz}$): Now suppose the original frequency was $292\text{ Hz}$. Adding wax will decrease this frequency. It will start moving down from $292\text{ Hz}$ towards the known $288\text{ Hz}$.

If the frequency drops by exactly $2\text{ Hz}$ due to the wax, the new frequency becomes $f_2^{\prime }=290\text{ Hz}$. Let's check the new beat frequency:

This perfectly matches the condition given in the problem! By starting at a higher frequency, the decrease caused by the wax brings the unknown fork closer to the known fork, thereby reducing the beat frequency from $4$ to $2$.

By systematically applying the physics of beats and the mechanics of harmonic oscillators, we have eliminated the incorrect path. The only logical conclusion is that the original frequency of the unknown tuning fork, before any wax was added, was indeed $292\text{ Hz}$.