A tuning fork of unknown frequency produces 3 | vedclass

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(A) The beat frequency is given by the formula: $	ext{beats} = |f_1 - f_2|$.Given, $f_2 = 384 \, 	ext{Hz}$ and $	ext{beat frequency} = 3 \, 	ext{H

Vedclass · Accepted Answer

$387$

Vedclass · Answer

(A) The beat frequency is given by the formula: $	ext{beats} = |f_1 - f_2|$.Given, $f_2 = 384 \, 	ext{Hz}$ and $	ext{beat frequency} = 3 \, 	ext{Hz}$.Therefore, $f_1 = 384 \pm 3$, which means $f_1 = 387 \, 	ext{Hz}$ or $f_1 = 381 \, 	ext{Hz}$.When wax is added to a tuning fork, its frequency $f_1$ decreases.It is given that the beat frequency decreases after adding wax.If $f_1 = 381 \, 	ext{Hz}$, adding wax would decrease $f_1$ further (e.g., to $380 \, 	ext{Hz}$), making the beat frequency $|380 - 384| = 4 \, 	ext{Hz}$, which is an increase.If $f_1 = 387 \, 	ext{Hz}$, adding wax would decrease $f_1$ (e.g., to $386 \, 	ext{Hz}$), making the beat frequency $|386 - 384| = 2 \, 	ext{Hz}$, which is a decrease.Since the beat frequency decreases, the original frequency must be $387 \, 	ext{Hz}$.

$A$ tuning fork of unknown frequency produces $3 \, \text{beats/sec}$ with a standard tuning fork of frequency $384 \, \text{Hz}$. The beat frequency decreases when a small piece of wax is placed on the prong of the unknown tuning fork. The frequency of the unknown tuning fork is .... $\text{Hz}$

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