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14.Waves and Sound
medium
A string is stretched between fixed points separated by $75.0\, cm$. It is observed to have resonant frequencies of $420\, Hz$ and $315\, Hz$. There are no other resonant frequencies between these two. Then, the lowest resonant frequency for this string is .... $Hz$
A
$105$
B
$1.05$
C
$1050$
D
$10.5$
(AIIMS-2008)
Solution
Given $\frac{\mathrm{nv}}{2 \ell}=315$ and $(\mathrm{n}+1) \frac{\mathrm{v}}{2 \ell}=420$
$\Rightarrow \frac{n+1}{n}=\frac{420}{315} \Rightarrow n=3$
Hence $3 \times \frac{\mathrm{v}}{2 \ell}=315 \Rightarrow \frac{\mathrm{v}}{2 \ell}=105 \mathrm{Hz}$
The lowest resonant frequency is when
$\mathrm{n}=1$
Therefore lowest resonant frequency
$=105 \mathrm{Hz}$
Standard 11
Physics
