Animated Solution for Physics - Waves: Two tuning forks with natural frequencies of $340\text{ Hz}$ each move relative to a stationary observer. One fork moves away from the observer, while the other moves towards him at the same speed. The observer hears beats of frequency $3\text{ Hz}$. Find the speed of the tuning fork. Speed of sound = $340\text{ m/s}$.

Imagine standing in a quiet open field. Suddenly, two identical tuning forks begin to vibrate nearby. One is rushing towards you, while the other is speeding away at the exact same rate. What you hear is not a steady, pure tone, but a rhythmic, pulsating throb—a rise and fall in loudness that we call beats. This is the beautiful intersection of the Doppler effect and wave interference.

When a source of sound moves relative to a stationary observer, the pitch we perceive changes. This phenomenon, known as the Doppler Effect, is a cornerstone of wave mechanics. As the source approaches, it chases its own sound waves, compressing them and increasing the frequency. Conversely, as it recedes, it leaves the waves stretched out behind it, lowering the frequency.

In our problem, we have two identical tuning forks with a natural frequency of $f=340\text{ Hz}$. One fork is moving towards the observer with speed $v_s$, while the other is moving away with the same speed $v_s$. The speed of sound in air is given as $v=340\text{ m/s}$.

A Tale of Compressed Waves

Let's write down the mathematical expressions for the frequencies heard by our stationary observer. For the approaching tuning fork, the sound waves are compressed. This results in a higher perceived frequency, which we denote as $f_1$:

For the receding tuning fork, the sound waves are stretched out. This results in a lower perceived frequency, which we denote as $f_2$:

Notice how the denominator for the approaching source has a minus sign, making the fraction larger than $1$, while the receding source has a plus sign, making the fraction smaller than $1$. This perfectly matches our physical intuition!

When these two waves of slightly different frequencies, $f_1$ and $f_2$, reach the observer's ears simultaneously, they superimpose. Because their frequencies are close but not identical, they periodically go in and out of phase. This leads to alternating constructive and destructive interference, which we perceive as a periodic variation in loudness.

The frequency of this pulsation is the Beat Frequency ($f_b$), which is simply the absolute difference between the two shifted frequencies:

We are given that the observer hears exactly $3$ beats per second. Therefore, our beat frequency is $f_b=3\text{ Hz}$.

Now, let's substitute our Doppler shift expressions into the beat frequency equation:

By factoring out the common term $f\cdot v$, we can simplify the expression inside the parentheses:

Finding a common denominator for the fractions inside the parentheses gives:

Simplifying the numerator, we get:

At this point, we could solve this equation exactly, but a clever physicist always looks for elegant approximations that simplify the math without losing accuracy. Notice that the beat frequency of $3\text{ Hz}$ is extremely small compared to the natural frequency of $340\text{ Hz}$. This tells us that the frequency shift is tiny, which in turn means the speed of the tuning forks, $v_s$, must be much smaller than the speed of sound, $v$ ($v_s\ll v$).

Since $v_s\ll v$, the term $v_s^2$ is completely negligible compared to $v^2$. Therefore, we can make the approximation:

Substituting this approximation back into our equation yields:

One of the $v$ terms in the numerator cancels with one in the denominator, leaving us with:

Now, let's substitute our known values: $f=340\text{ Hz}$ and $v=340\text{ m/s}$. Watch how beautifully the numbers align:

The $340$ in the numerator and denominator cancel out completely! This leaves us with an incredibly simple linear equation:

Solving for $v_s$, we find:

Thus, the speed of each tuning fork is exactly $1.5\text{ m/s}$.

Approximate: A Physicist's Rigor

To appreciate the power of our approximation, let's solve the exact quadratic equation without neglecting $v_s^2$:

Using the quadratic formula to solve for $v_s$, we get:

The difference between the exact answer ($1.49999\text{ m/s}$) and our approximate answer ($1.5\text{ m/s}$) is less than $0.001\%$! This demonstrates how powerful and reliable physical approximations can be when used correctly.