Question

Two spherical bob of masses $M_A$ and $M_B$ are hung vertically from two strings of length $l_A$ and $l_B$ respectively. They are executing SHM with frequency relation $f_A=2f_B$, Then:

• A. $l_A=\frac {l_B}{4}$
• B. $l_A=4l_B$
• C. $l_A=2l_B$ & $M_A=2M_B$
• D. $l_A=\frac {l_B}{2}$ & $M_A=\frac {M_B}{2}$

Answer 1

Answer: (A)

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

Answer 2

$f=\frac {1}{2\pi} \sqrt {\frac gl}$

$f∝\frac {1}{\sqrt l}$

$\frac {f_A}{f_B} =\sqrt {\frac {l_B}{l_A}}$

⇒ $\frac {2f_B}{f_B} =\sqrt {\frac {l_B}{l_A}}$

⇒ $4=\frac {l_B}{l_A}$

⇒ $l_A=\frac {l_B}{4}$

So, the correct option is (A): $l_A=\frac {l_B}{4}$