A block ok is connected to a spring such that its time period under normal condition would be T full stop block is compressed by a distance A and released. And Elastic wall is located in front of block at a distance of A/2 the time period of oscillation of the block.
Answers
Answer:
When the block is compressed by a distance A and released, it will start to oscillate back and forth. As it approaches the elastic wall located at a distance of A/2, the wall will exert a force on the block, causing it to slow down and change direction.
Explanation:
To calculate the time period of oscillation of the block after it hits the elastic wall, we need to consider the effect of the wall on the block's motion. The wall will act as a point of reflection, causing the block to change direction and start moving in the opposite direction. This change in direction will cause the block's velocity to change sign, which will affect the time period of oscillation.
To calculate the new time period, we need to use the formula:
T' = 2π√(m/k')
where T' is the new time period, m is the mass of the block, and k' is the effective spring constant after the block hits the elastic wall.
To calculate k', we can use the formula:
k' = k/(1 + 2L/A)
where k is the spring constant of the spring, L is the distance between the block and the wall before compression, and A is the distance that the block is compressed. Using these formulas, we can calculate the time period of oscillation of the block after it hits the elastic wall.
To know more about the concept please go through the links:
https://brainly.in/question/33571177
https://brainly.in/question/6246756
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