Question

Two strings $A$ and $B$ have lengths $i_A$ and $l_B$ and carry pendulum of masses $M_A$ and $M_B$ at their lower ends the upper ends being supported by rigid supports. If $n_A$ and $n_B$ are their frequencies of their oscillations and $n_A=2n_B$, then :

• A. $l_A=4l_B$, regardless of masses
• B. $l_B=4l_A$, regardless of masses
• C. $M_A=2M_B,l_A=2l_B$
• D. $M_B=2M_A,l_B=2l_A$

Answer 1

The correct option is B $l_B=4l_A$, regardless of masses

$n_A=\frac{1}{2\pi }\sqrt{\frac{g}{l_A}}$

$n_B=\frac{1}{2\pi }\sqrt{\frac{g}{l_B}}$

$\because n_A=2n_B$

$\Rightarrow \frac{1}{2\pi }\sqrt{\frac{g}{l_A}}=2\times \frac{1}{2\pi }\sqrt{\frac{g}{l_B}}$

$\Rightarrow \frac{1}{l_A}=\frac{4}{l_B}$

$\Rightarrow l_B=4l_A$

Hence,

option $\left(B\right)$ is correct answer.