**Problem 2.59: Undamped Single-Degree-of-Freedom System**An undamped single-degree-of-freedom system consists of a mass \(5 \, \text{kg}\) and a spring of stiffness \(2{,}000 \, \text{N/m}\). Find the *response* of the system when the mass is subjected to the following initial conditions:(a) \( x_0 = 20 \, (\text{mm}) \), \( \dot{x}_0 = 200 \, \left( \frac{\text{mm}}{\text{sec}} \right) \)(b) \( x_0 = -20 \, (\text{mm}) \), \( \dot{x}_0 = 200 \, \left( \frac{\text{mm}}{\text{sec}} \right) \)(c) \( x_0 = 20 \, (\text{mm}) \), \( \dot{x}_0 = -200 \, \left( \frac{\text{mm}}{\text{sec}} \right) \)(d) \( x_0 = -20 \, (\text{mm}) \), \( \dot{x}_0 = -200 \, \left( \frac{\text{mm}}{\text{sec}} \right) \)

Introduction In this problem, we will calculate the system's response at the initial condition's different values. We use the motion equation of the single degree of freedom undamped vibration to calculate the system response. Given data: massm=5 kgspring stifnessk=2000 N/m Find out: The response of the system when the mass is subjected to the following initial conditions: (a). X0=20 mm, Xo˙=200 mm/sec (b). X0=-20 mm, Xo˙=200 mm/sec (c). X0=20 mm, Xo˙=-200 mm/sec (d). X0=-20 mm, Xo˙=-200 mm/secCalculation of the response of the system when the mass is subjected to the following initial condition. (a). X0=20 mm, (Xo˙)=200 mm/sec For the single-degree undamped free vibration, the expression of the equation of motion is given below, mx¨+kx=0m(x¨+kxm)=0x¨+kxm=0D2+kmy=0 (D operator method)D2+km=0D=±kmiX(t)=c1coskm t+c2sinkm tkm=2000 N/m2 kgkm=20 rad/sX(t)=c1coskm t+c2sinkm tX(t)= c1cos(20t)+c2sin(20t) .............equation(1) c1 and c2 can be calculated by substituting the initial condition in equation (1). X(t)=c1cos(20t)+c2sin(20t)Xo=20 mmat, t=0X(0)= c1cos(20t)+c2sin(20t)20=c1cos20×0+c2sin(20×0)20=c1×1+0c1=20 mmX(t)= c1cos(20t)+c2sin(20t)X˙(t)= c1-sin(20t)20+c2cos(20t)20X˙(t)=200 mm/secat, t=0X˙(0)=200 mm/sec200= 20×-sin(20×0)20+c2cos(20×0)20200=0+c2×1×20200=20c2c2=10 mm By substituting the value of c1 and c2 in equation (1) we get the response of the system. X(t)= c1cos(20t)+c2sin(20t) c1=20 mmc2=10 mmX(t)=c1cos(20t)+c2sin(20t) X(t)=20×cos(20t)+10×sin(20t) Hence the response of the system X(t)=20×cos(20t)+10×sin(20t)The response of the system for the given initial conditions, (b). X0=-20 mm, (Xo˙)=200 mm/sec X(t)=c1cos(20t)+c2sin(20t)Xo=-20 mmat, t=0X(0)=c1cos(20×0)+c2sin(20×0)-20=c1×1+0c1=-20 mmX(t)= c1cos(20t)+c2sin(20t)X˙(t)= c1-sin(20t)20+c2cos(20t)20X˙(t)=200 mm/secat, t=0X˙(0)=200 mm/sec200= 20×-sin(20×0)20+c2cos(20×0)20200=0+c2×1×20200=20c2c2=10 mm By substituting the value of c1 and c2 in equation (1) we get the response of the system. X(t)= c1cos(20t)+c2sin(20t) c1=-20 mmc2=10 mmX(t)=c1cos(20t)+c2sin(20t) X(t)=-20×cos(20t)+10×sin(20t) Hence the response of the system X(t)=-20×cos(20t)+10×sin(20t)The response of the system for the given initial conditions, (c). X0=20 mm, (Xo˙)=-200 mm/sec X(t)=c1cos(20t)+c2sin(20t)Xo=20 mmat, t=0X(0)=c1cos(20×0)+c2sin(20×0)20=c1×1+0c1=20 mmX(t)= c1cos(20t)+c2sin(20t)X˙(t)= c1-sin(20t)20+c2cos(20t)20X˙(t)=-200 mm/secat, t=0X˙(0)=-200 mm/sec-200= 20×-sin(20×0)20+c2cos(20×0)20-200=0+c2×1×20-200=20c2c2=-10 mm By substituting the value of c1 and c2 in equation (1) we get the response of the system. X(t)= c1cos(20t)+c2sin(20t) c1=20 mmc2=-10 mmX(t)=c1cos(20t)+c2sin(20t) X(t)=20×cos(20t)-10×sin(20t) Hence the response of the system X(t)=20×cos(20t)-10×sin(20t)The response of the system for the given initial conditions, (d). X0=-20 mm, (Xo˙)=-200 mm/sec X(t)=c1cos(20t)+c2sin(20t)Xo=-20 mmat, t=0X(0)=c1cos(20×0)+c2sin(20×0)-20=c1×1+0c1=-20 mmX(t)= c1cos(20t)+c2sin(20t)X˙(t)= c1-sin(20t)20+c2cos(20t)20X˙(t)=-200 mm/secat, t=0X˙(0)=-200 mm/sec-200= 20×-sin(20×0)20+c2cos(20×0)20-200=0+c2×1×20-200=20c2c2=-10 mm By substituting the value of c1 and c2 in equation (1) we get the response of the system. X(t)= c1cos(20t)+c2sin(20t) c1=-20 mmc2=-10 mmX(t)=c1cos(20t)+c2sin(20t) X(t)=-20×cos(20t)-10×sin(20t) Hence the response of the system X(t)=-20×cos(20t)-10×sin(20t) Results: The system's response at the different initial conditions is obtained, (a). X(t)=20×cos(20t)+10×sin(20t) (b). X(t)=-20×cos(20t)+10×sin(20t) (c). X(t)=20×cos(20t)-10×sin(20t) (d). X(t)=-20×cos(20t)-10×sin(20t)