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A massless rod of length $L$ is suspended by two identical strings AB and CD of equal length. A block of mass $4 \mathrm{~m}$ is suspended from point O such that BO is equal to $(x-6)$ . Further it is observed that the frequency of $1^{\text {st }}$ harmonic in AB is equal to $2^{\text {nd }}$ harmonic frequency in CD x.
Option: 1
$\frac{L+20}{5}$

Option: 2
$\frac{L+30}{7}$

Option: 3
$\frac{L+30}{5}$

Option: 4
$\frac{L+40}{5}$

Answers (1)

best_answer

Frequency of the first harmonic of $AB=\frac{1}{2 l} \sqrt{\frac{T A B}{4 m}}$ . When $l$ is the length of the two strings.

Frequency of the $2^{\text {nd }}$ harmonic of CD

$=\frac{1}{l} \sqrt{\frac{T C D}{4 m}}$

Given that the two frequency are equal

$\frac{1}{2 l} \sqrt{\frac{T A B}{4 m}}=\frac{1}{l} \sqrt{\frac{T C D}{4 m}} \\$

$\frac{T A B}{4}=T C D \\$

$TAB=4TCD$

For rotational equilibrium of massless rod taking torque about Point O.

$TAB\times(x-6)={TCD}[L-(n-6)] \\$

$TAB \times(n-6)=\frac{T A B}{4}[L-(n-6)] \\$

$4(n-6)=L-(n-6) \\$

$4n-24=L-n+6 \\$

$4 x+x-24-6=L \\$

$5 x-30=L \\$

$5x=L+30$

$x=\frac{L+30}{5}$

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