Question

A tuning fork of frequency $512\, Hz$ makes $4$ beats per second with the vibrating string of a piano. The beat frequency decreases to $2$ beats per sec when the tension in the piano string is slightly increased. The frequency of the piano string before increasing the tension was

• A. 510 Hz
• B. 514 Hz
• C. 516 Hz
• D. 508 Hz

Answer 1

The Correct Option is D

To solve this problem, let's first understand the concept of beat frequency. When two sound waves of different frequencies are played together, they create a phenomenon known as beats. The beat frequency is calculated as the absolute difference between the two frequencies:

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

Here, f_1 is the frequency of the tuning fork, which is 512 Hz, and f_2 is the frequency of the piano string that we need to determine.

Step 1: Initial Condition

The tuning fork makes 4 beats per second initially. This yields the equation:

|512 - f_2| = 4

Solving for f_2, we get two possible solutions:

• f_2 = 512 + 4 = 516\, \text{Hz}
• f_2 = 512 - 4 = 508\, \text{Hz}

Step 2: Condition After Increasing Tension

When the tension is increased slightly, the beat frequency decreases to 2 beats per second. The frequency of the string increases with increased tension (based on the physics of strings), so the new frequency will be closer to the tuning fork frequency, reducing the beat frequency.

The new equation with the decreased beat frequency is:

|f' - 512| = 2

This indicates that the new string frequency f' is either:

• f' = 512 + 2 = 514\, \text{Hz}
• f' = 512 - 2 = 510\, \text{Hz}

Since increasing tension causes the frequency to increase, the older frequency must be less than 512 Hz. Thus, the correct original frequency is 508 Hz.

Conclusion:

The frequency of the piano string before increasing the tension was 508 Hz, which matches the correct choice amongst the options provided.