## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 10 - Problems - Page 403: 97

#### Answer

$y(t) = (1.6~cm)~cos~[~(24.75~rad/s)~t~]$

#### Work Step by Step

In general: $y(t) = A~cos(\omega~t+\phi)$ Since the body starts at the highest position, there is no phase shift, so $\phi = 0$ We can find the amplitude. Note that the amplitude is the distance between the highest point and the equilibrium position. At the equilibrium position, the upward force of the spring is equal in magnitude to the weight: $k~A = mg$ $A = \frac{mg}{k}$ $A = \frac{4.0~N}{250~N/m}$ $A = 0.016~m$ $A = 1.6~cm$ We can find $\omega$: $\omega = \sqrt{\frac{k}{m}}$ $\omega = \sqrt{\frac{k}{(4.0~N/g)}}$ $\omega = \sqrt{\frac{k~g}{4.0~N}}$ $\omega = \sqrt{\frac{(250~N/m)(9.80~m/s^2)}{4.0~N}}$ $\omega = 24.75~rad/s$ We can write an equation to describe the motion of the body: $y(t) = (1.6~cm)~cos~[~(24.75~rad/s)~t~]$

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