A 9.00 kg mass is connected to a spring with a spring constant of 4.00 N/m.The velocity is given by the expression v(t) = 10.0 cm/s sin($\omega$ t)+ 8.00 cm/s cos($\omega$ t).What is the maximum displacement of the simple harmonic motion?
A) 25.1 cm
B) 30.5 cm
C) 35.3 cm
D) 40.4 cm
E) 45.5 cm

Question

Quizplus · Accepted Answer

The maximum displacement (amplitude) can be found using the formula $ A = \sqrt{v_0^2 + u_0^2} / \omega $, where $ v_0 = 10.0 $ cm/s, $ u_0 = 8.00 $ cm/s, and $ \omega = \sqrt{k/m} = \sqrt{4/9} $ rad/s. Calculating gives $ A = \sqrt{10^2 + 8^2} / \sqrt{4/9} = 25.1 $ cm.

A 900 Kg Mass Is Connected to a Spring with a a Spring