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

510 Hz

514 Hz

516 Hz

508 Hz

Solution

508 Hz

suppose space straight n subscript straight p space equals space frequency space of space piano space equals space ? space left parenthesis straight n subscript straight p space proportional to space square root of straight T straight n subscript straight f space equals space frequency space of space tuning space fork space equals space 512 space Hz straight x space equals space beat space frequency space equals space 4 space beats divided by straight s. space which space is space decreasing space left parenthesis 4 space rightwards arrow 2 right parenthesis space after changing space the space tension space of space piano space wire space is space increasing space so space straight n subscript straight p space upwards arrow Hence comma space straight n subscript straight p space upwards arrow space minus space straight n subscript straight f space equals space straight x downwards arrow space rightwards arrow space wrong straight n subscript straight f space minus space straight n subscript straight p space upwards arrow space equals space straight x space downwards arrow space rightwards arrow space correct rightwards double arrow space straight n subscript straight p space equals space straight n subscript straight f space minus space straight x space equals space 512 minus 4 space equals space 508 space Hz

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