Wave Mechanics Question 592
Question: Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of the system is
Options:
A) $ \frac{1}{2\pi }\sqrt{\frac{k}{4M}} $
B) $ \frac{1}{2\pi }\sqrt{\frac{4k}{M}} $
C) $ \frac{1}{2\pi }\sqrt{\frac{k}{7M}} $
D) $ \frac{1}{2\pi }\sqrt{\frac{7k}{M}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] Springs on the left of the block are in series, hence their equivalent spring constant is $ K _{1}=\frac{(2k)(2k)}{2k+2k}=k $ Springs on the right of the block are in parallel, hence their equivalent spring constant is $ k _{2}=k+2k=3k $ Now again both $ K _{1} $ and $ K _{2} $ are in parallel
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Therefore ,K _{eq}=k _{1}+k _{2}=k+3k=4k $ Hence, frequency is $ f=\frac{1}{2\pi }\sqrt{\frac{K _{eq}}{M}}=\frac{1}{2\pi }\sqrt{\frac{4k}{M}} $
BETA
