Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A tuning fork of frequency $512 \text{ Hz}$ makes $4 \text{ beats/s}$ with the vibrating strings of a piano. The beat frequency decreases to $2 \text{ beats/s}$ when the tension in the piano strings is slightly increased. The frequency of the piano string before increasing the tension was:

$510 \text{ Hz}$

$514 \text{ Hz}$

$516 \text{ Hz}$

$508 \text{ Hz}$

Step-by-Step Solution

Determine possible initial frequencies: The frequency of the tuning fork is $f = 512 \text{ Hz}$ . The beat frequency is $4 \text{ beats/s}$ . Therefore, the initial frequency of the piano string ( $f_p$ ) can be either $512 + 4 = 516 \text{ Hz}$ or $512 - 4 = 508 \text{ Hz}$ .
Analyze the effect of tension: When the tension in a string increases, its fundamental frequency increases because frequency is directly proportional to the square root of tension ( $f \propto \sqrt{T}$ ) .
Determine the correct frequency:

If the initial frequency was $516 \text{ Hz}$ , increasing the tension would increase the frequency further (e.g., to $518 \text{ Hz}$ ). The new beat frequency with the $512 \text{ Hz}$ tuning fork would be $|518 - 512| = 6 \text{ beats/s}$ , which represents an increase in the beat frequency.
If the initial frequency was $508 \text{ Hz}$ , increasing the tension would increase the frequency closer to $512 \text{ Hz}$ (e.g., to $510 \text{ Hz}$ ). The new beat frequency would be $|510 - 512| = 2 \text{ beats/s}$ , which is a decrease and matches the condition given in the problem. Therefore, the original frequency of the piano string was $508 \text{ Hz}$ .

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