Engineering vibrations 4th edition solutions

Next let the spring experience a dynamic deflection x(t) governed by summing the forces in (b) to get (a) (b) From a force balance in the static case (a): mg = kxs, where xs is the static deflection of the spring. X ( t ) = Asin (! nt + ' ) = 1sin ( 5t + 2 ) andДЇigure P1.5 Solution: The free-body diagram of problem system in (a) for the static case and in (b) for the dynamic case, where x is now measured from the static equilibrium position. In particular, using static equilibrium along with Newtons law, determine what effect gravity has on the equation of motion and the systems natural frequency. ! v ( t ) = x ( t ) = A! n cos (! nt + ' ) = 5 cos ( 5t + 2 ) Setting t = 0 in these expressions yields: x(0) = 1sin(2) = 0.9093 mm v(0) = 5 cos(2) = - 2.081 mm/s1.5 Find the equation of motion for the hanging spring-mass system of Figure P1.5, and compute the natural frequency. Solution: Given: A = 1 mm, ' = 2 rad, ! = 5 rad/s. Assume the response is of the form x(t) = Asin(! nt + ' ). Calculate the initial conditions that caused this vibration to occur. The phase shift from t = 0 is measured to be 2 rad and the frequency is found to be 5 rad/s. The amplitude of vibration of an undamped system is measured to be 1 mm. ' This is identical to problem 2, except v0 = 0. X Solve m + kx = 0 for k = 4 N/m, m = 1 kg, x0 = 1 mm, and v0 = 0. Combining the two underlined expressions (2 eqs in 2 unkowns): '2c1 + 2 ' 2c1 = 5 i ! c1 = 1 5 1 5 ' i, and c2 = + i 2 4 2 4 Substitute initial conditions: x0 = 1 mm, v0 = K = 2 rad/s m The sum of x1 and x2 is also a solution so that the total solution is: where ( n =x = x1 + x 2 = c1e 2it + c 2 e !2it Thus there are two solutions:x1 = c1e! k $ # m i& t ' % Mr 2 + k = 0 r= k i m! k $ # ' m i& t ' % Substitute into equation (1) to x get: mar 2 ert + kaert = 0rt Solution: Given: (1) m!! + kx = 0 x rt 2 rt ! Assume: x(t) = ae. X Derive the solution of m + kx = 0 and plot the result for at least two periods for the case with n = 2 rad/s, x0 = 1 mm, and v0 = 5 mm/s. The sample standard deviation in computed stiffness is:m 15 m(kg) x(m) Solution: Free-body diagram:kx k m mg20ДЇrom the free-body diagram and static equilibrium: Plot the data and calculate the spring's stiffness. Problems and Solutions Section 1.1 (1.1 through 1.19) 1.1 The spring of Figure 1.2 is successively loaded with mass and the corresponding (static) displacement is recorded below.

Engineering vibrations 4th edition solutions