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A tuning fork of freqqency 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 [2010]
  • a)
    510 Hz
  • b)
    514 Hz
  • c)
    516 Hz
  • d)
    508 Hz
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
A tuning fork of freqqency 512 Hz makes 4 beats per second with the vi...
The frequency of the pianostring = 512 ± 4 = 516 or 508. When the tension is increased, beat frequency decreases to 2, it means that frequency of the string is 508 as frequency of string increases with tension.
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Most Upvoted Answer
A tuning fork of freqqency 512 Hz makes 4 beats per second with the vi...
To solve this question, we need to understand the concept of beat frequency and its relation to the difference in frequencies between two vibrating sources.

- The beat frequency is the difference between the frequencies of two sources that are producing sound waves. It is perceived as a periodic variation in loudness or intensity of sound.
- When two sources have frequencies close to each other, they produce beats. The number of beats per second is equal to the difference in frequencies between the two sources.
- In this question, we are given that a tuning fork of frequency 512 Hz makes 4 beats per second with the vibrating string of a piano.

Let's break down the given information:

1. The tuning fork frequency is 512 Hz.
2. The tuning fork and the piano string produce 4 beats per second.

We are asked to find the frequency of the piano string before increasing the tension.

To find the answer, we need to understand how the beat frequency changes when the tension in the piano string is increased:

- When the tension in the piano string is slightly increased, the beat frequency decreases to 2 beats per second.
- This means the difference in frequencies between the tuning fork and the piano string has decreased.

Now, let's solve the problem step by step:

Step 1: Find the initial difference in frequencies between the tuning fork and the piano string:
- Since the tuning fork has a frequency of 512 Hz, the initial frequency of the piano string can be calculated by subtracting the beat frequency from the tuning fork frequency.
- Initial difference in frequencies = 512 Hz - 4 beats per second = 512 Hz - 4 Hz = 508 Hz

Step 2: Find the final difference in frequencies between the tuning fork and the piano string:
- Since the beat frequency decreases to 2 beats per second when the tension is increased, the final difference in frequencies can be calculated by subtracting the beat frequency from the tuning fork frequency.
- Final difference in frequencies = 512 Hz - 2 beats per second = 512 Hz - 2 Hz = 510 Hz

Step 3: Find the frequency of the piano string before increasing the tension:
- The frequency of the piano string before increasing the tension is equal to the tuning fork frequency minus the final difference in frequencies.
- Frequency of the piano string = 512 Hz - 510 Hz = 2 Hz

Therefore, the frequency of the piano string before increasing the tension was 508 Hz, which corresponds to option D.
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A tuning fork of freqqency 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 [2010]a)510 Hzb)514 Hzc)516 Hzd)508 HzCorrect answer is option 'D'. Can you explain this answer?
Question Description
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