In the given figure,two elastic rods A and B | vedclass

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Question

(A) The motion of the mass $m$ consists of two parts: free motion between the rods and the time spent during collisions with the rods.$1$. Time taken

Vedclass · Accepted Answer

$\frac{2L}{v} + 2\pi \sqrt{\frac{mL}{AY}}$

Vedclass · Answer

(A) The motion of the mass $m$ consists of two parts: free motion between the rods and the time spent during collisions with the rods.$1$. Time taken to travel the distance $L$ between the rods: The mass travels distance $L$ twice in one cycle (once to the right and once to the left). Thus,$t_{free} = \frac{2L}{v}$.$2$. Time spent in collisions: The rods act as springs. For rod $B$ (area $A$,modulus $Y$,length $L$),the spring constant is $k_B = \frac{YA}{L}$. The time spent in contact with rod $B$ is half the time period of a spring-mass system: $t_B = \frac{1}{2} 	imes 2\pi \sqrt{\frac{m}{k_B}} = \pi \sqrt{\frac{mL}{AY}}$.$3$. For rod $A$ (area $A/2$,modulus $2Y$,length $L$),the spring constant is $k_A = \frac{(2Y)(A/2)}{L} = \frac{YA}{L}$. The time spent in contact with rod $A$ is $t_A = \frac{1}{2} 	imes 2\pi \sqrt{\frac{m}{k_A}} = \pi \sqrt{\frac{mL}{AY}}$.$4$. Total time period $T = t_{free} + t_A + t_B = \frac{2L}{v} + \pi \sqrt{\frac{mL}{AY}} + \pi \sqrt{\frac{mL}{AY}} = \frac{2L}{v} + 2\pi \sqrt{\frac{mL}{AY}}$.

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