Question 531: Waves

Question

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 1st harmonic (fundamental frequency) in AB is equal to 2 nd harmonic frequency in CD. Then, length of BO is

Accepted Answer

Frequency of $1^{	ext {st }}$ harmonic of $\mathrm{AB}=\frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}}$
Frequency of $2^{	ext {nd }}$ harmonic of $\mathrm{CD}=\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}}$
Given that 2 frequencies are equal.
$$ \begin{aligned} \frac{1}{2 l} \sqrt{\frac{\mathrm{~T}_{\mathrm{AB}}}{m}} & =\frac{1}{l} \sqrt{\frac{\mathrm{~T}_{\mathrm{CD}}}{m}} \ \frac{\mathrm{~T}_{\mathrm{AB}}}{m} 	imes \frac{1}{4 l^2} & =\frac{1}{l^2} \cdot \frac{\mathrm{~T}_{\mathrm{CD}}}{m} \quad(	ext { On squaring }) \ \frac{\mathrm{T}_{\mathrm{AB}}}{4} & =\mathrm{T}_{\mathrm{CD}} \ \mathrm{~T}_{\mathrm{AB}} & =4 \mathrm{~T}_{\mathrm{CD}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $$
For rotational equilibrium of massless rod, Taking torque about point D ,
$$ \mathrm{T}_{\mathrm{AB}} 	imes x=\mathrm{T}_{\mathrm{CD}}(\mathrm{~L}-x) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$$
For translational equillibrium,
$$ \mathrm{T}_{\mathrm{AB}}+\mathrm{T}_{\mathrm{CD}}=m g \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$$
From (i) and (iii), $4 \mathrm{~T}_{\mathrm{CD}}+\mathrm{T}_{\mathrm{CO}}=m g$
$$ \begin{array}{ll} \Rightarrow & \mathrm{T}_{\mathrm{CD}}=\frac{m g}{5} \ 	herefore & \mathrm{~T}_{\mathrm{AB}}=4 	imes \frac{m g}{5}=\frac{4 m g}{5} \end{array} $$
On substituting above value in (i) we get
$$ \Rightarrow \quad \begin{aligned} \frac{4 m g}{5} 	imes x & =\frac{m g}{5}(L-x) \ 4 x & =\mathrm{L}-x \ 5 x & =\mathrm{L} \ x & =\mathrm{L} / 5 . \end{aligned} $$

Answer

$\frac{\mathrm{L}}{5}$

Answer

$\frac{4 \mathrm{~L}}{5}$

Answer

$\frac{3 \mathrm{~L}}{4}$

Answer

$\frac{\mathrm{L}}{4}$

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