A (0.500-mathrm{kg}) block oscillates up and down on a vertical spring for which (k=12.5 mathrm{~N} / mathrm{m}).
Question:
A \(0.500-\mathrm{kg}\) block oscillates up and down on a vertical spring for which \(k=12.5 \mathrm{~N} / \mathrm{m}\). The initial amplitude of the motion is \(0.100 \mathrm{~m}\).
(a) What is the natural angular frequency \(\omega\) ?
(b) If the observed angular frequency is \(\omega_{\mathrm{d}}=4.58 \mathrm{~s}^{-1}\), what is \(b\) ?
(c) Write an equation for the position as a function of time.
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a The natural angular frequency omega of the oscillation can be found using the formula omega sqrtfr...View the full answer
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