Question

Masses $M_A$ and $M_B$ hanging from the ends of strings of lengths $L_A$ and $L_B$ are executing simple harmonic motions. If their frequencies are $f_A=2f_B$, then

• A. $L_A=2L_B$ and $M_A=M_B/2$
• B. $L_A=4L_B$ regardless of masses
• C. $L_A=L_B/4$ regardless of masses
• D. $L_A=2L_B$ and $M_A=2M_B$

Answer 1

Answer: (C)

$f_A=\frac{1}{2\pi }\sqrt{\frac{g}{L_A}}$
and $f_B=\frac{f_A}{2}=\frac{1}{2\pi }\sqrt{\frac{g}{L_B}}$
$\therefore \frac{f_A}{f_{A/2}}=\frac{1}{2\pi }\sqrt{\frac{g}{L_A}}\times 2\pi \sqrt{\frac{L_B}{g}}$
$\Rightarrow 2=\sqrt{\frac{L_B}{L_A}}$
$\Rightarrow 4=\frac{L_B}{L_A}$, regardless of mass.