A free-spring system, damped and not forced hung downwards has the following data: mass = 6 kg, d...
A free-spring system, damped and not forced hung downwards has the following data: mass = 6 kg, damping coefficient of 6 newtons / meter / sec and coefficient of spring (3/2) newton / meter. If it is stretched down 0.6 meters, and fanned upward with a speed of 0.6 m / sec, determine the time it takes to cross to the equilibrium position for the first time. Answer:
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A free-spring system, damped and not forced hung downwards has the following data: mass = 6 kg, d...
5) A damped simple harmonic oscillator consists of a.40 kg mass oscillating vertically on a spring...
5) A damped simple harmonic oscillator consists of a.40 kg mass oscillating vertically on a spring with k- 15 N/m with a damping coefficient of .20 kg/s. The spring is initially stretched 17 cm downwards and the mass is released from rest. a) What is the angular frequency of the mass? b) What is the position of the mass at t-3 seconds? c) Sketch a position vs time graph for the mass, showing at least 5 full cycles of oscillation....
Solve it with matlab 25.16 The motion of a damped spring-mass system (Fig. P25.16) is described...
Solve it with matlab
25.16 The motion of a damped spring-mass system (Fig. P25.16) is described by the following ordinary differential equation: d’x dx ++ kx = 0 m dr dt where x = displacement from equilibrium position (m), t = time (s), m 20-kg mass, and c = the damping coefficient (N · s/m). The damping coefficient c takes on three values of 5 (under- damped), 40 (critically damped), and 200 (overdamped). The spring constant k = 20 N/m....
I want matlab code. 585 i1 FIGURE P22.15 22.15 The motion of a damped spring-mass system (Fig. P22.15) is de...
I want matlab code.
585 i1 FIGURE P22.15 22.15 The motion of a damped spring-mass system (Fig. P22.15) is described by the following ordinary differ- ential equation: dx dx in dt2 dt where x displacement from equilibrium position (m), t time (s), m 20-kg mass, and c the damping coefficient (N s/m). The damping coefficient c takes on three values of 5 (underdamped), 40 (critically damped), and 200 (over- damped). The spring constant k-20 N/m. The initial ve- locity is...
a-slug mass is hung onto a spring, whereupon the spring is stretched 6 in from its...
a-slug mass is hung onto a spring, whereupon the spring is stretched 6 in from its natural length. The mass is n started in motion from the equilibrium position with an initial velocity of 4 ft/sec in the upward direction. Find the subsequent motion of the mass, if the force due to air resistance is -2t lb.
(1 point) A frictionless spring with a 7-kg mass can be held stretched 0.6 meters beyond...
(1 point) A frictionless spring with a 7-kg mass can be held stretched 0.6 meters beyond its natural length by a force of 70 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1 m/sec, find the position of the mass after t seconds. 0.244sin(4.08t) meters
2. A spring is stretched 10 cm by a force of 3 newtons. A mass of...
2. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 newtons when the velocity of the mass is 5 m/sec. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/sec, determine its position u at time t. Find the quasi frequency and...
(1 point) A spring-mass system with a 5-kg mass and a damping constant 8-N sec/m can...
(1 point) A spring-mass system with a 5-kg mass and a damping constant 8-N sec/m can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value y2 – 4mk? Find the position of the mass after t seconds. Your answer should be a function of the variable t...
A block of mass 0.427 kg is hung from a vertical spring and allowed to reach...
A block of mass 0.427 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result the spring is stretched by 0.612 m. Find the spring constant Number N/m The block is then pulled down an additional 0.317 m and released from rest. Assuming no damping, what is its period of oscillation? Number How high above the point of release does the block reach as it oscillates? Number In
Differential Equation problem We know that a force of 2.8 Newtons is required to stretch a certain spring 0.7 meters beyond its natural length. A 1.44-kg mass is attached to this spring and allowed t...
Differential Equation problem
We know that a force of 2.8 Newtons is required to stretch a certain spring 0.7 meters beyond its natural length. A 1.44-kg mass is attached to this spring and allowed to come to equilibrium. The mass-spring system is then set in motion by applying a push in the upward direction that gives the mass an initial velocity of 1.04 meters per second. Let y(t) represent the displacement of the mass above the equilibrium position t seconds...
Spring mass damper system with forced response, the forced system given by the equation For damping...
Spring mass damper system with forced response, the forced system given by the equation For damping factor:E-0.1 ; mass; m-| kg: stiffness of spring; k-100 Nm; f-| 00 N; ω Zun; initial condition: x (0)-2 cms; r(0) = 0. fsincot Task Marks 1. Write down the reduced equation into 2first orderns Hand written equations differential equations 2. Rearrange equation (1) with the following generalized equation 250, x+osinor calculations 3. Calculate the value of c calculations Hand calculations 4. Using the...
5) A damped simple harmonic oscillator consists of a.40 kg mass oscillating vertically on a spring with k- 15 N/m with a damping coefficient of .20 kg/s. The spring is initially stretched 17 cm downwards and the mass is released from rest. a) What is the angular frequency of the mass? b) What is the position of the mass at t-3 seconds? c) Sketch a position vs time graph for the mass, showing at least 5 full cycles of oscillation....
Solve it with matlab
25.16 The motion of a damped spring-mass system (Fig. P25.16) is described by the following ordinary differential equation: d’x dx ++ kx = 0 m dr dt where x = displacement from equilibrium position (m), t = time (s), m 20-kg mass, and c = the damping coefficient (N · s/m). The damping coefficient c takes on three values of 5 (under- damped), 40 (critically damped), and 200 (overdamped). The spring constant k = 20 N/m....
I want matlab code.
585 i1 FIGURE P22.15 22.15 The motion of a damped spring-mass system (Fig. P22.15) is described by the following ordinary differ- ential equation: dx dx in dt2 dt where x displacement from equilibrium position (m), t time (s), m 20-kg mass, and c the damping coefficient (N s/m). The damping coefficient c takes on three values of 5 (underdamped), 40 (critically damped), and 200 (over- damped). The spring constant k-20 N/m. The initial ve- locity is...
a-slug mass is hung onto a spring, whereupon the spring is stretched 6 in from its natural length. The mass is n started in motion from the equilibrium position with an initial velocity of 4 ft/sec in the upward direction. Find the subsequent motion of the mass, if the force due to air resistance is -2t lb.
(1 point) A frictionless spring with a 7-kg mass can be held stretched 0.6 meters beyond its natural length by a force of 70 newtons. If the spring begins at its equilibrium position, but a push gives it an initial velocity of 1 m/sec, find the position of the mass after t seconds. 0.244sin(4.08t) meters
2. A spring is stretched 10 cm by a force of 3 newtons. A mass of 2 kg is hung from the spring and is also attached to a viscous damper that exerts a force of 3 newtons when the velocity of the mass is 5 m/sec. If the mass is pulled down 5 cm below its equilibrium position and given an initial downward velocity of 10 cm/sec, determine its position u at time t. Find the quasi frequency and...
(1 point) A spring-mass system with a 5-kg mass and a damping constant 8-N sec/m can be held stretched 0.5 meters beyond its natural length by a force of 2.5 newtons. Suppose the spring is stretched 1 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value y2 – 4mk? Find the position of the mass after t seconds. Your answer should be a function of the variable t...
A block of mass 0.427 kg is hung from a vertical spring and allowed to reach equilibrium at rest. As a result the spring is stretched by 0.612 m. Find the spring constant Number N/m The block is then pulled down an additional 0.317 m and released from rest. Assuming no damping, what is its period of oscillation? Number How high above the point of release does the block reach as it oscillates? Number In
Differential Equation problem
We know that a force of 2.8 Newtons is required to stretch a certain spring 0.7 meters beyond its natural length. A 1.44-kg mass is attached to this spring and allowed to come to equilibrium. The mass-spring system is then set in motion by applying a push in the upward direction that gives the mass an initial velocity of 1.04 meters per second. Let y(t) represent the displacement of the mass above the equilibrium position t seconds...
Spring mass damper system with forced response, the forced system given by the equation For damping factor:E-0.1 ; mass; m-| kg: stiffness of spring; k-100 Nm; f-| 00 N; ω Zun; initial condition: x (0)-2 cms; r(0) = 0. fsincot Task Marks 1. Write down the reduced equation into 2first orderns Hand written equations differential equations 2. Rearrange equation (1) with the following generalized equation 250, x+osinor calculations 3. Calculate the value of c calculations Hand calculations 4. Using the...
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