Asked by Anonymous
A block of mass m=.75kg is fastened to an unstrained horizontal springwhose spring constant is 82 N/m. The block is given a displacement of +.12m, where + sign indicates that the displacement is along the +x axis, and then released from rest.
Find the angular frequency w of the resulting oscillatory motion.
For this question, I got 10.646. Is this correct?
What is the maximum speed of the block?
Would I need to find the amplitude to multiply by w to get the maximum speed? If that's the case, how would I find the amplitude?
Thank you!
Find the angular frequency w of the resulting oscillatory motion.
For this question, I got 10.646. Is this correct?
What is the maximum speed of the block?
Would I need to find the amplitude to multiply by w to get the maximum speed? If that's the case, how would I find the amplitude?
Thank you!
Answers
Answered by
drwls
The angular frequency w is sqrt (k/m). sqrt (82/.75) = 10.456 sec^-1
The maximum speed V can be obtained from the stored potential energy is the spring before release.
(1/2) k X^2 = (1/2) m V^2
V = sqrt (k/m) X = w X
where X is the 0.12 m displacement
The maximum speed V can be obtained from the stored potential energy is the spring before release.
(1/2) k X^2 = (1/2) m V^2
V = sqrt (k/m) X = w X
where X is the 0.12 m displacement
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