A massless rod of length is suspended by two identical strings and of equal length. A block of mass is suspended from point such that is equal to . Further it is observed that the frequency of 1st harmonic in is equal to 2nd harmonic frequency in .

Question

A massless rod of length $L$ is suspended by two identical strings $A B$ and $C D$ of equal length. A block of mass $m$ is suspended from point $O$ such that $B O$ is equal to $x$ . Further it is observed that the frequency of 1st harmonic in $A B$ is equal to 2nd harmonic frequency in $C D$ . $‘ x ’$ is [2006]

Smart Achievers · Accepted Answer

(1)

$Frequency of first harmonic in AB = f_{1} = \frac{1}{2 l} \sqrt{\frac{T}{μ}}$

$A B$

$f_{1} = f_{2} (given)$

$∴$ $\frac{1}{2 l} \sqrt{\frac{T_{1}}{μ}} = \frac{1}{l} \sqrt{\frac{T_{2}}{μ}}$ $or$ $T_{1} = 4 T_{2} . . . (i)$

Equating torques due to $T_{1}$ and $T_{2}$ about $O$ for rotational equilibrium,

$∴$ $T_{1} x = T_{2} (L - x)$

For translational equilibrium,

$T_{1} + T_{2} = m g$

From eq. (i) and (iii),

$T_{1} = \frac{4 m g}{5} and T_{2} = \frac{m g}{5}$

Now from eq. (ii),

$\frac{4 m g}{5} \times x = \frac{m g}{5} (L - x)$ $\Rightarrow 4 x = L - x$

$∴$ $x = \frac{L}{5}$

Solution

1 $\frac{L}{5}$

2 $\frac{4 L}{5}$

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

4 $\frac{L}{4}$

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Solution

Q. A massless rod of length L 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 in AB is equal to 2nd harmonic frequency in CD. ‘x’ is [2006]

1 L5

2 4L5

3 3L4

4 L4

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1 $\frac{L}{5}$

2 $\frac{4 L}{5}$

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

4 $\frac{L}{4}$