PART A k = 2.000 N/m M=0.125kg A 0.125kg mass is attached to a horizontal spring of spring constant k = 2 N/m and set into oscillation. Assume frictionless floor. 4 rodls What is the angular frequency o of the vibration? @ = How about the vibrating frequency f in Hz? f = 0.64 Hz The position of the mass is given as x = A cos (@t + o0). With your calculated value of w, fill in the following table (don't forget to set your calculator to "radian mode"). Each row represents a different oscillation with the same spring, pay attention to sign: v (m/s) .092 - .14 a (m/s²) 3076 -32 -.8712 -.64 x (m) -.02 A (m) t (s) po (rad) F (N) .04 -.04 V 0.03 2.5 0.3 -212 1.36 0.04 0.02 0.2 0.06 .0545 0.1 0.4 ,076 1.08 -. 08 0.04 0.3 -26 - 25 -0.2 2.94 0.5 0.04 0.1 0.1 PART B Each of the rows in the following table represents an oscillation of a different spring and mass. Fill in the blanks. E is the total energy. Vmax, v, Xmax, and x are all positive in values. E (J) (rad/s) A (m) Vmax Xmax m (kg) k (N/m) x at t=0 v at t=0 (m/s) (m/s²) (m) (m/s) 0.05 20 0.01 1.0 9.0 0.3 0.5 0.15 2.0 5.0 2.0 5.0 0.6 2.0 0.6 0.2 2.0 5.0 0.1

PART A k = 2.000 N/m M=0.125kg A 0.125kg mass is attached to a horizontal spring of spring constant k = 2 N/m and set into oscillation. Assume frictionless floor. 4 rodls What is the angular frequency o of the vibration? @ = How about the vibrating frequency f in Hz? f = 0.64 Hz The position of the mass is given as x = A cos (@t + o0). With your calculated value of w, fill in the following table (don't forget to set your calculator to "radian mode"). Each row represents a different oscillation with the same spring, pay attention to sign: v (m/s) .092 - .14 a (m/s²) 3076 -32 -.8712 -.64 x (m) -.02 A (m) t (s) po (rad) F (N) .04 -.04 V 0.03 2.5 0.3 -212 1.36 0.04 0.02 0.2 0.06 .0545 0.1 0.4 ,076 1.08 -. 08 0.04 0.3 -26 - 25 -0.2 2.94 0.5 0.04 0.1 0.1 PART B Each of the rows in the following table represents an oscillation of a different spring and mass. Fill in the blanks. E is the total energy. Vmax, v, Xmax, and x are all positive in values. E (J) (rad/s) A (m) Vmax Xmax m (kg) k (N/m) x at t=0 v at t=0 (m/s) (m/s²) (m) (m/s) 0.05 20 0.01 1.0 9.0 0.3 0.5 0.15 2.0 5.0 2.0 5.0 0.6 2.0 0.6 0.2 2.0 5.0 0.1

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

See similar textbooks

Similar questions

1 -) an undamaged small inverted pendulum given in the figure on the sideinfer the expression that gives the natural frequency of its vibrations.Admissions that you make, etc. please type. Rod rigid andit's massless. Static equilibrium position of the systemas can be seen, it is in a vertical direction. 2-) only one of the springs in the figure (right or left) if it was replaced by a damper with a damper coefficient chow would it affect the natural frequency? Required statementsextract and interpret.
1. A system is needed to be designed as damped spring mass system. Calculate the amplitude of vibration. Stiffness=10 kN/m; mass 2Mg; and dashpot coefficient 2 kNs/m. It is subjected to a harmonic force = 500N @ frequency 0.5Hz ANS. X₁-43mm 2. What is the A length of three springs connected in series and parallel? Change in length of the spring is 4 cm. reelle k₁ llll пише m=160 gram
k, k3 A В k2 What is the equivalent spring stiffness between the points A and B. If ki = 1 N/m, k2 =1 N/m. k3 = 2 N/m and k4 = 2 N/m A N/m
Q2: Find the E.O.M of the system shown in the fig by using newtons law, and then find the total forced displecment if yo = 0.001m and the time of one period is 0.125 s and w₁ = 45.25 rad / sec body 1.6 m/s +
1- Two mass-spring-damper systems A & B. System A has = 0.1 and wn = 20 rad/sec, and system B has < = 0.3 and w₁ = 5 rad/sec. Which system will vibrate longer in time? Why?
1- Two mass-spring-damper systems A & B. System A has = 0.1 and @n = 20 rad/sec, and system B has < = 0.3 and wn = 5 rad/sec. Which system will vibrate longer in time? Why?
Given the vibration system with multiple degree of freedom. For string 1, k1 = 10kN/m %3D For string 2, k2 = 20KN/m For string 3, k3 = 30 kN/m %3D mass = 500kg mass of the beam will not be consider. Solve for the following: i. Frequency j. Period
Consider the following spring system. Assume down is the positive direction. Write the stiffness mirix K- • Compute the displacements caused by the external forces f = - Displacement = C₂ m mi heees3 m₂ Hom 3 7/////// with spring constants c = 134
B-increase of spring coefficient like 0.5 k, 2k, 5k and / orhow does the natural frequency of the system decreasereferencesexternal links
F(1) Spring m. Ball Dashpot Find the motion of the mass-spring system with the driving force F(t) shown above for the given data. [Hint: Read Forced Oscillations notes.] e-2t, 0 2 y(0) = 1, y'(0) =-1.5.
A wagon weighing 5100 Kg moving at a speed of 8 km/h has to be brought to rest. Springs made of wire diameter 25mm with a Dm of 250mm and with 24 turns are available. Find the number of springs required in buffer to stop the wagon at a compression of 180mm. b)A close coiled helical spring is made of diameter wire 25mm. The spring index is 8. Findthe number of turns required and maximum allowable load if the allowable shear stress is100 MPa and elongation is limited to 40mm. [G=80 GPa
PROBLEM A 1) Find the equivalent spring, N/mm 2) Find the force, F k. Kl: =7 K5. =11 0000 K2 =8 X-10 mm k К3 -9 КА -10 К6 -12 ....... ........ .....