Question

The frequency of tuning fork \(\mathrm{A}\) is \(2 \%\) more than the frequency of a standard fork. Frequency of tuning fork \(B\) is \(3 \%\) less than the frequency of the standard fork. If 6 beats per second are heard when the two forks \(\mathrm{A}\) and \(\mathrm{B}\) are excited, then frequency of \(A\) is \(\ldots \ldots \ldots . H z\) (A) 120 (B) \(122.4\) (C) \(116.4\) (D) 130

Short answer

The frequency of tuning fork A is \(122.4\, Hz\), which corresponds to option (B).

Step-by-step solution

Setting up the equations

Let the standard frequency be S. According to the problem, frequency of A is 2% more than S, so: \(A = S + 0.02S = 1.02S\) Similarly, frequency of B is 3% less than S, so: \(B = S - 0.03S = 0.97S\) We also know that the absolute difference between the frequencies of A and B is 6 Hz. \(|A - B| = 6\)

Solving for the standard frequency (S)

We know that \(A = 1.02S\) and \(B = 0.97S\). Substitute these into the equation \(|A - B| = 6\), we get: \(|1.02S - 0.97S| = 6\) Solve for S: \(|0.05S| = 6\) \(S = \frac{6}{0.05}\) \(S = 120\)

Find the frequency of A

We now know the standard frequency is 120 Hz. To find the frequency of A, we substitute the value of S into the equation: \(A = 1.02S\) \(A = 1.02 \times 120\) \(A = 122.4\, Hz\) Therefore, the frequency of tuning fork A is 122.4 Hz, which corresponds to option (B).

Key concepts

Beat Frequency

When two sound waves of different frequencies are played together, they interfere with each other (a process called superposition). This results in a fluctuation or variation in loudness, which we perceive as beats.
The beat frequency is simply the number of these fluctuations or cycles in loudness per second. It can be calculated using the formula:

• Beat Frequency = \(|f_1 - f_2|\)

Here, \(f_1\) and \(f_2\) are the frequencies of the two sound sources. In our problem, these are tuning forks A and B.
It's stated that 6 beats per second are heard when tuning forks A and B are excited past.
This implies that the beat frequency is 6 Hz, which means \(|A - B| = 6\). This helps us in setting up an equation to solve for the unknown frequencies.

Percent Change in Frequency

Understanding percent change is crucial in problems dealing with frequency adjustments. A percent change describes how much a value increases or decreases in terms of a percentage of the original value.
For tuning fork A, its frequency is 2% more than a standard frequency S. This means:

• Tuning Fork A Frequency = Standard Frequency + 2% of Standard Frequency.
• \(A = S + 0.02S = 1.02S\)

Similarly, for tuning fork B, which is 3% less than S:

• Tuning Fork B Frequency = Standard Frequency - 3% of Standard Frequency.
• \(B = S - 0.03S = 0.97S\)

By understanding these percentage changes, we can set up the equations necessary to solve for the unknown frequencies. It's a straightforward application of what percent increase and percent decrease mean.

Equation Solving

To find the unknown frequencies, setting up the correct equations is key. Based on the problem, we have two expressions:

• \(A = 1.02S\)
• \(B = 0.97S\)

And we also know that \(|A - B| = 6\). The equation becomes:

• \(|1.02S - 0.97S| = 6\)
• This simplifies to \(|0.05S| = 6\)

We can solve for S (the standard frequency) by dividing both sides by 0.05:

• \(S = \frac{6}{0.05}\)
• \(S = 120\)

Once we have S, calculating the frequency of tuning fork A is straightforward:

• \(A = 1.02 \times 120\)
• \(A = 122.4\, Hz\)

This logical progression from setting up an equation through solving it is a clear example of how mathematical reasoning can lead to a solution. It's all about breaking down the problem into manageable steps and dealing with them one at a time.