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14.Waves and Sound
normal
A massless rod is suspended by two identical strings $AB$ and $CD$ of equal length. A block of mass $m$ is suspended from point $ O $ such that $BO$ is equal to $’x’$. Further, it is observed that the frequency of $1^{st}$ harmonic (fundamental frequency) in $AB$ is equal to $2^{nd}$ harmonic frequency in $CD$. Then, length of $BO$ is
A
$\frac{L}{5}$
B
$\frac{L}{4}$
C
$\frac{4L}{5}$
D
$\frac{3L}{4}$
Solution
$\mathrm{n}_{1}=\mathrm{n}_{2}$
$\frac{1}{{2l}}\sqrt {\frac{{{{\rm{T}}_1}}}{{\rm{m}}}} = \frac{2}{{2l}}\sqrt {\frac{{{{\rm{T}}_2}}}{{\rm{m}}}} $
$\mathrm{T}_{\mathrm{1}}=4 \mathrm{T}_{2}$
$\tau_{1}=\tau_{2}$
$\mathrm{T}_{1} \times \mathrm{x}=\tau_{2} \times(\mathrm{L}-\mathrm{x})$
$4{\rm{x}} = l – {\rm{x}}$
${\rm{x}} = \frac{l}{5}$
Standard 11
Physics
