Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 15 - Oscillations - Problems - Page 442: 103a

Answer

$T = 0.44~s$

Work Step by Step

We can use the original period to find the mass of the block: $T = 2\pi \sqrt{\frac{m}{k}}$ $T^2 = \frac{4\pi^2 m}{k}$ $m = \frac{T^2~k}{4\pi^2}$ $m = \frac{(0.40~s)^2 (600~N/m)}{4\pi^2}$ $m = 2.43~kg$ We can find the new period: $T = 2\pi \sqrt{\frac{m}{k}}$ $T = 2\pi \sqrt{\frac{2.43~kg+0.50~kg}{600~N/m}}$ $T = 0.44~s$
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