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A tuning fork vibrating with a sonometer having $20\,cm$ wire produces $5$ beats per sec. The beat frequency does not change if the length of the wire is changed to $21\,cm$. The frequency of the tuning fork must be:
A. 200 Hz
B. 210 Hz
C. 205 Hz
D. 215 Hz

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Last updated date: 13th Jul 2024
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Answer
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Hint: In order to solve this question, you must be aware about the relationship between the frequency of the tuning fork and the length of the wire. Frequency is the number of occurrences of a repeating event per unit time.

Complete step by step answer:
Let frequency of the tuning fork be $n$. Now, the possible frequencies of the sonometer are $(n + 5)$ and $(n - 5)$. As the frequency of the vibration string $ \propto $ $\dfrac{1}{{length}}$. Therefore, for wire of length $20\,cm$, frequency must be $(n + 5)$ and for wire of length $21\,cm$, the frequency must be $(n - 5)$.
$n + 5 = \dfrac{1}{{2 \times 20}}\sqrt {\dfrac{T}{m}} $ ……. (1)
$\Rightarrow n - 5 = \dfrac{1}{{2 \times 21}}\sqrt {\dfrac{T}{m}} $ ……... (2)
Dividing eq. (1) by (2), we get
$\dfrac{{n + 5}}{{n - 5}} = \dfrac{{21}}{{20}}$
$\therefore n = 205$ Hz

Hence, option C is correct.

Note:The four main factors affecting frequency are length, diameter, tension and density. When any of these factors of a string are changed, it will vibrate with a different frequency. Shorter strings have higher frequencies and therefore higher pitch.