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posted by  boboknight on 11/7/2009 9:44:09 AM  |  status: Closed  |  Earned Karma: 50

Spring

Course Textbook Chapter Problem Needs by
Calculus Based Physics Principles of Physics (3rd) by Serway, Jewett N/A N/A 11/12/2009 at 8:00:00 PM
Question Details:
A 0.500kg mass oscillates in simple harmonic motion at the end of a spring of constant 150N/m. The total energy of the system is 2.50J. What is the amplitude of the motion (in m)?

Hint:

For a mass on a spring: E = (1/2) k (x(max))^2 = (1/2) m (v(max))^2
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posted by watt (MNK) on 11/7/2009 9:52:38 AM  |  status: Live
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Response Details:
mass m = 0.5 kg
Spring constant  k = 150 N / m
Total energy  E = 2.5 J
we know  E = ( 1/ 2) k A ^ 2
from this amplitude of the motion  A = √[ 2E / k ]
                                                       = 0.1825 m
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