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A block of mass `2kg` executes simple harmonic motion under the reading from at aspring .The angular and the time period of motion are `0.2 cm` and `2pi`sec respectively Find the maximum force execut by the spring in the block.A. `0.05 N`B. `0.002 N`C. `0.003 N`D. `0.004 N` |
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Answer» Correct Answer - D The mass of block `= m = 2kg amp = A = 0.2cm` and time period `T = 6.28 sec` The maximum force exerted will be `KA` and it will be at extreme position. `f_(max) = m omega^(2) A [omega = sqrt((K)/(m)) "where" K = "spring constant"]` `= 2 xx ((2pi)/(6.28))^(2) xx (0.2 xx 10^(-2)) = 0.004 N` Acceleration at `0.1 cm` is given by Acceleration `a= - omega^(2)x = - ((2pi)/(6.28))^(2) xx (0.1cm)` ` = -0.1 cm//sec = - 10^(-3) m//s^(2)` |
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