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 $M_A=M_B / 2$

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(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.