| Two identical carts A and B each with mass m are connected via a spring with spring constant k. Two additional springs, identical to the first, connect the carts to two fixed points. The carts are free to oscillate under the effect of the springs in one dimensional frictionless motion. |
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| Under suitable initial conditions, the two carts will oscillate in phase according to \[{{x}_{A}}(t){{x}_{0}}\sin {{\omega }_{1}}t={{x}_{B}}(t)\] Where \[{{x}_{A}}\] and \[{{x}_{B}}\] are the locations of carts A and B relative to their respective equilibrium positions. Under other suitable initial conditions, the two carts will oscillate exactly out of phase according to \[{{x}_{A}}(t)={{x}_{0}}\sin {{\omega }_{2}}t=-{{x}_{B}}(t)\] Determine the ratio\[{{\omega }_{2}}/{{\omega }_{1}}\]. |
A) \[\sqrt{3}\]
B) 2
C) \[2\sqrt{2}\]
D) 3
Correct Answer: A
Solution :
[a] Under first condition middle spring is relaxed all the time only \[L-R\]spring make. So \[{{\omega }_{1}}=\sqrt{\frac{k}{m}}\] In second condition mid pt. of middle spring is at permanent rest and left hand of middle spring and left spring can be considered in parallel and the effective constant will be \[K=2K=3K\] So \[{{\omega }_{2}}=\sqrt{\frac{3K}{m}}\]
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