%3Cp%3EA%200.420-kg%20object%20attached%20to%20a%20spring%20with%20a%20force%20c
ID: 2295494 • Letter: #
Question
%3Cp%3EA%200.420-kg%20object%20attached%20to%20a%20spring%20with%20a%20force%20constant%20of%0A8.00%20N%2Fm%20vibrates%20in%20simple%20harmonic%20motion%20with%20an%20amplitude%20of%0A11.2%20cm.%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(a)%20Calculate%20the%20maximum%20value%20of%20its%20speed.%20Incorrect%3A%20Your%0Aanswer%20is%20incorrect.%20cm%2Fs%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(b)%20Calculate%20the%20maximum%20value%20of%20its%20acceleration.%0Acm%2Fs2%20%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(c)%20Calculate%20the%20value%20of%20its%20speed%20when%20the%20object%20is%209.20%20cm%0Afrom%20the%20equilibrium%20position.%20cm%2Fs%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(d)%20Calculate%20the%20value%20of%20its%20acceleration%20when%20the%20object%20is%0A9.20%20cm%20from%20the%20equilibrium%20position.%20cm%2Fs2%26nbsp%3B%3C%2Fp%3E%0A%3Cp%3E%3Cbr%20%2F%3E%3C%2Fp%3E%0A%3Cp%3E(e)%20Calculate%20the%20time%20interval%20required%20for%20the%20object%20to%20move%0Afrom%20x%20%3D%200%20to%20x%20%3D%203.20%20cm.%20s%3C%2Fp%3E%0AExplanation / Answer
w = angular frequency = k/m = 8 / .420 = 19 Hz
max speed = Aw =
A = amplitude = 11.2 x 19 = 213.333 cm/s
max accelration = Aw^2 = 4043.2 cm/s^2
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