1 secA block-spring system has a maximum restoring force Fmax = 0.1 N. If theamplitude of the motion is A = 0.16 m and the mass of the block is m = 100 g thenthe angular frequency w is equal toO 1.25 rad/s20 rad/s5 rad/s2.5 rad/sO 10 rad/sA simple pendulum is set into motion at two different times with different initialconditions, The first time: the bob is in its equilibrium position and is given an A simple pendulum is set into motion at two different times with different initialconditions. The first time: the bob is in its equilibrium position and is given aninitial velocity +4 m/s. The maximum height reached by the bob is h1,max and theperiod of motion is T1. The second time: the bob is in its equilibrium position andis given an initial velocity +2 m/s. The maximum height reached by the bob ish2,max and the period of motion is T2. Assume that the motion is simpleharmonic, which of the following is true?h2.max = 4h1max and T, = 27,h = 4hmax and T, = T%3D%3D2,maxThen we have:Then we have:

A block-spring system has a maximum restoring force Fmax = 0.1 N. If the amplitude of the motion is A = 0.16 m and the mass of the block ism% D 100 g then the angular frequency w is equal to O 1.25 rad/s 20 rad/s 5 rad/s O 2.5 rad/s O 10 rad/s

A block-spring system has a maximum restoring force Fmax = 0.1 N. If the amplitude of the motion is A = 0.16 m and the mass of the block ism% D 100 g then the angular frequency w is equal to O 1.25 rad/s 20 rad/s 5 rad/s O 2.5 rad/s O 10 rad/s

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Consider a mass-spring system shown below. X1 m k₁ ooooo The system model is given by d²x m m₁ 1 d1² dx, 2 di² 1 · + (k₁₂ + k₂) x ₂ −k ₂x₂=F₁ (1) 1 k₂ ooooo F₁ k₂x₂ + (k₂ + k ₂) x₂ = F₂(t) k₂ +k z 2 m_s (m_s²³ + k₁+k₂ ²+k₁ + k ₂ ) ( m ₂s ² + k₂ + k ₂) -k 2 where F₁(t) and F₂(t) are the inputs to the system. Find the transfer function X₁(s)/F₁(s). k ₂ (m₁s²+k₁+k₂) (m₂s² + k ₂2 +k₂ ) − k²/ 2 ₁s² + k ₁+k₂) 1 (m₂x²+₁+k₂) (m₂x²+k₂+^₂)=R} 2 ) ( m²₂ s² + k₂ + k ₂ 1 2 m₂ k ₁+k₂ 1 (m₁s²+k₁ + k₂) (m₂s² + k₂ + k 3) − k ²/ k 3 (m₂x²+k₂+k₂) (m₂x²+k₂+k₂) −R} k ² (m₂s² + k ₂+k₂) (m₂x²+k₂ +k₂) (m₂x²+k₂+A₂)-R² 3) k3 ooooo F₂
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