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Step-by-Step Solution
Step 1: Understand the Resonance Condition for a String
When a string fixed at both ends vibrates in normal modes (or harmonics), the
$n^{th}$ resonant frequency is given by
$ \frac{nv}{2\ell} $,
where
• $n$ is a positive integer (the mode number),
• $v$ is the wave speed on the string, and
• $\ell$ is the length of the string (distance between the two fixed ends).
Step 2: Relate the Consecutive Resonant Frequencies
We are told that the string has two consecutive resonant frequencies: $315\,Hz$ and $420\,Hz$, with no mode in between.
Therefore, these correspond to mode numbers $n$ and $n+1$ in the formula
$ \frac{nv}{2\ell} $. Thus,
$ \frac{nv}{2\ell} = 315 $
and
$ \frac{(n+1)\,v}{2\ell} = 420 $.
Step 3: Use the Ratio to Find the Mode Number
Dividing the second frequency by the first:
$ \frac{\frac{(n+1)\,v}{2\ell}}{\frac{n\,v}{2\ell}} = \frac{420}{315} $.
Simplifying,
$ \frac{n+1}{n} = \frac{420}{315} = \frac{4}{3} $.
Hence,
$ n+1 = \frac{4}{3}\,n \ \Rightarrow\ 3(n+1) = 4n \ \Rightarrow\ 3n + 3 = 4n \ \Rightarrow\ n = 3 $.
Step 4: Determine the Wave Speed Constant $\frac{v}{2\ell}$
From $ \frac{3v}{2\ell} = 315 $, we get
$ \frac{v}{2\ell} = \frac{315}{3} = 105\,Hz $.
Step 5: Find the Lowest Resonant Frequency
The lowest resonant frequency (the fundamental mode) corresponds to $n=1$, giving
$ \frac{1 \cdot v}{2\ell} = 105\,Hz $.
Hence, the string's lowest resonant frequency is $105\,Hz$.
Final Answer
The lowest resonant frequency of the string is $105\,Hz$.
