# sol13 - HW Solutions 13 8.01 MIT Prof Kowalski Harmonic...

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HW Solutions #1 3 - 8.01 MIT - Prof. Kowalski Harmonic Oscillators and Relative Motion . 1 ) Vibration Isolation System a ) Let’s denote the table displacement from equilibrium with Y . The Newton’s second law f y = ma y for the box will read: M d 2 Y dt 2 = k ( y −Y )( 1 ) Where y denotes the displacement of ceiling from the situation that table is in equilibrium. The form of y is given: y = A cos ωt Replace this in (1) you’ll get: M d 2 Y dt 2 = k Y + kA cos Which is equivalent to an additional force - other than the spring - cos acting on the table. b ) Substitute Y = C ( ω )cos in (1) and divide two sides by M: ω 2 C ( ω = ω 2 0 ( A C ( ω )) cos Where I used the fact that k M = ω 2 0 . This gives: C ( ω )= A 1 ω 2 ω 2 0 c ) Set ω 0 2 π Hz and ω =15 × 2 π Hz C ( ω ) A = 1 1 ω 2 ω 2 0 1 1 15 2 ∼− 0 . 005 So the amplitude is reduced by a factor 200. 1

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2 ) 13.88 A Rod’s Oscillation via Spring Please refer to the figure 13.88. There is a horizontal force which acts at the end of the metal rod which points to the left and it comes from the stretched spring. The magnitude of this force for
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sol13 - HW Solutions 13 8.01 MIT Prof Kowalski Harmonic...

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