Oscillations - Result Question 51
55. 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 $f_A=$ $2 f_B$, then
(a) $L_A=2 L_B$ and $M_A=M_B / 2$
[2000]
(b) $L_A=4 L_B$ regardless of masses
(c) $L_A=L_B / 4$ regardless of masses
(d) $L_A=2 L_B$ and $M_A=2 M_B$
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Answer:
Correct Answer: 55. (c)
Solution:
(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.
