Wave Mechanics Question 590
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
Options:
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} $
Show Answer
Answer:
Correct Answer: 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.
BETA
