answer:To find the displacement of the mass after 2 seconds, we can use the equation for simple harmonic motion: x(t) = A * cos(ωt + φ) where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and φ is the phase shift. First, we need to determine the angular frequency ω. This can be found using the formula: ω = √(k/m) where k is the spring constant (10 N/m) and m is the mass (0.5 kg). ω = √(10/0.5) = √20 = 2√2 rad/s Next, we need to find the amplitude A. Since the mass is initially displaced by 0.1 m, we have: A = 0.1 m Now, we need to find the phase shift φ. Since the mass is initially at its maximum displacement, the phase shift is π/2 (90°). Finally, we can find the displacement after 2 seconds: x(2) = A * cos(ω * 2 + φ) x(2) = 0.1 * cos(2 * 2√2 + π / 2) x(2) = 0.1 * cos(4√2 + π / 2) x(2) ≈ 0.1 * cos(6.2832 - 1.5708) x(2) ≈ 0.1 * cos(4.7124) x(2) ≈ 0.1 * (-0.4062) x(2) ≈ -0.0406 m So, the displacement of the mass after 2 seconds is approximately -0.0406 m. The negative sign indicates that the mass has traveled to the left of its equilibrium position.,To find the displacement of the mass after 2 seconds, we can use the equation for the displacement of a mass-spring system without damping: x(t) = A * cos(ω * t + φ) where x(t) is the displacement at time t, A is the amplitude of oscillation, ω is the angular frequency, and φ is the phase angle. First, we need to find the angular frequency ω. This can be calculated using the formula: ω = sqrt(k/m) where k is the spring constant (10 N/m) and m is the mass (0.5 kg). ω = sqrt(10/0.5) = sqrt(20) ≈ 4.47 rad/s Since the mass is initially displaced by 0.1 m and then released, the amplitude A is 0.1 m, and the phase angle φ is 0 (because the mass starts at its maximum displacement). Now we can find the displacement after 2 seconds: x(2) = 0.1 * cos(4.47 * 2 + 0) ≈ 0.1 * cos(8.94) ≈ -0.062 m So the displacement of the mass after 2 seconds is approximately -0.062 m, which means it is 0.062 m to the left of the equilibrium position., We have a mass-spring system with no damping. We're given the spring constant (k = 10 N/m), initial displacement (x_0 = 0.1 m), and the mass (m = 0.5 kg). We want to find the displacement of the mass after 2 seconds. First, we need to find the angular frequency of the system: ω = sqrt(k/m) ω = sqrt(10 N/m / 0.5 kg) ω = sqrt(20 rad^2/s^2) ω = 4.472 rad/s Since there is no damping, the motion of the mass will be simple harmonic motion. For this type of motion, we can use the formula for displacement as a function of time: x(t) = A * cos(ωt - φ) where A is the amplitude, ω is the angular frequency, t is time, and φ is the phase angle. However, since we don't have information about the phase angle, we cannot find the exact displacement at a specific time. But we can find the amplitude of the motion using the initial conditions: x_0 = A * cos( - φ) Since the mass is initially displaced by 0.1 m, we have: 0.1 m = A * cos( - φ) Since the cosine function has a maximum value of 1 and a minimum value of -1, the amplitude A must be greater than or equal to |x_0|: A >= |x_0| A >= 0.1 m Therefore, the displacement of the mass after 2 seconds will be in the range: 0.1 m <= x(2 s) <= sqrt(20) m 0.1 m <= x(2 s) <= 4.472 m Without knowing the phase angle, we cannot determine the exact displacement at t = 2 s, but we know it will be within this range.

answer:To find the position function x(t) for the mass-spring system, we need to solve the differential equation for the system and apply the given initial conditions. The differential equation for a mass-spring system with no damping is: m * x''(t) + k * x(t) = 0 where m is the mass of the object (0.5 kg in this case), x(t) is the position of the object as a function of time, k is the spring constant (10 N/m in this case), and x''(t) is the second derivative of the position function with respect to time. We can first rewrite the equation in terms of the angular frequency ω = sqrt(k/m) as: x''(t) + ω^2 * x(t) = 0 Next, we can use the method of undetermined coefficients to find a solution to this differential equation. The general solution to the equation is: x(t) = A * cos(ω * t) + B * sin(ω * t) where A and B are constants that can be determined from the initial conditions. We are given the initial displacement x(0) = 2 cm or x(0) = 0.02 m, and since the system starts from rest, we also have x'(0) = 0 m/s. We can use these initial conditions to determine the values of A and B. Differentiating x(t) with respect to time, we get: x'(t) = -A * ω * sin(ω * t) + B * ω * cos(ω * t) Now, applying the initial conditions: x(0) = A * cos(0) + B * sin(0) = 0.02 x'(0) = -A * ω * sin(0) + B * ω * cos(0) = 0 From the first equation, we get A = 0.02, and from the second equation, since sin(0) is 0, we get B = 0. So, the position function for the mass-spring system with the given initial conditions is: x(t) = 0.02 * cos(ω * t) where ω is the angular frequency of the system, given by ω = sqrt(k/m) = sqrt(10/0.5) = 4.47 rad/s.,To find the position function x(t) for this mass-spring system, we need to solve the differential equation for a simple harmonic oscillator. The differential equation for the position function x(t) is: m * x''(t) + c * x'(t) + k * x(t) = 0 where m is the mass, c is the damping coefficient, x'(t) is the first derivative of x(t) with respect to time (equal to the velocity), and x''(t) is the second derivative of x(t) with respect to time (equal to the acceleration). In this problem, we are given: m = 0.5 kg c = 0 (since the system is undamped) k = 10 N/m x(0) = 0.02 m (displaced 2 cm from equilibrium) x'(0) = 0 (released from rest) The equation becomes: 0.5 * x''(t) + 0 * x'(t) + 10 * x(t) = 0 This is a second-order linear homogeneous differential equation with constant coefficients, which has the general solution: x(t) = A * cos(ω * t) + B * sin(ω * t) where A and B are constants, and ω is the angular frequency: ω = sqrt(k/m) Plugging in the given values: ω = sqrt(10/0.5) = sqrt(20) To find the constants A and B, we use the initial conditions: x(0) = 0.02 x'(0) = 0 Applying the first initial condition: 0.02 = A * cos(0) + B * sin(0) Since sin(0) = cos(0) = 0, we get: 0.02 = A Applying the second initial condition: 0 = -A * ω * sin(0) + B * ω * cos(0) Simplifying: 0 = B * ω Since A ≠ 0, we have ω ≠ 0, so: B = 0,To solve this problem, we will first write down the differential equation for the mass-spring system and then use the initial conditions to find the position function x(t). The differential equation for a mass-spring system without damping is given by: m * x''(t) + k * x(t) = 0 where m is the mass, k is the spring constant, and x(t) is the position function of the mass. In our case, m = 0.5 kg and k = 10 N/m. 0.5 * x''(t) + 10 * x(t) = 0 Now, we need to solve this second-order linear homogeneous differential equation. We can guess a solution of the form x(t) = e^(rt), where r is a constant. Plugging this into the differential equation, we get: 0.5 * r^2 * e^(rt) + 10 * e^(rt) = 0 Dividing by e^(rt), we get the characteristic equation: 0.5 * r^2 + 10 = 0 Solving for r, we get two complex roots: r1, r2 = ±(i * sqrt(20)) The general solution for x(t) is given by: x(t) = A * cos(sqrt(20) * t) + B * sin(sqrt(20) * t) Now, we need to use the initial conditions to find the constants A and B. The initial conditions are: x(0) = 0.02 m (2 cm displacement from equilibrium) x'(0) = 0 m/s (mass is released from rest) Applying the initial conditions: x(0) = A * cos(0) + B * sin(0) = A A = 0.02 m To find B, we need to differentiate x(t) with respect to time: x'(t) = -A * sqrt(20) * sin(sqrt(20) * t) + B * sqrt(20) * cos(sqrt(20) * t) Applying the second initial condition: x'(0) = -A * sqrt(20) * sin(0) + B * sqrt(20) * cos(0) = B * sqrt(20) B = 0 m/s So, the position function x(t) is: x(t) = 0.02 * cos(sqrt(20) * t)

answer:The equation of motion for a simple harmonic oscillator, like the mass-spring system, is given by: x(t) = A * cos(ωt + φ) where x(t) is the position of the mass at time t, A is the amplitude, ω is the angular frequency, and φ is the phase angle. First, we need to find the angular frequency (ω) using the formula ω = sqrt(k/m), where k is the spring constant and m is the mass: ω = sqrt(16 N/m / 2 kg) = sqrt(8 rad/s^2) = 2.8284 rad/s Now, we can find the amplitude (A) using the initial displacement: A = 0.3 meters Since the initial velocity is not zero, there will be a non-zero phase angle (φ). We can find it using the formula v(0) = -A*ω*sin(φ), where v(0) is the initial velocity: 0.2 m/s = -0.3 m * 2.8284 rad/s * sin(φ) sin(φ) = 0.2 m/s / (0.3 m * 2.8284 rad/s) sin(φ) ≈ 0.2357 φ ≈ 13.93 degrees So, the equation of motion of the mass is: x(t) = 0.3 * cos(2.8284t + 13.93°) The frequency (f) of oscillations can be found using the formula f = ω / (2*π): f = 2.8284 rad/s / (2*π) ≈ 0.4461 Hz The period (T) of oscillations can be found using the formula T = 1/f: T = 1/0.4461 Hz ≈ 2.242 seconds,To determine the equation of motion for the mass, we can use the following formula for a simple harmonic oscillator: x(t) = A * cos(ωt + φ) where x(t) is the displacement of the mass at time t, A is the amplitude of the oscillation, ω is the angular frequency, and φ is the phase angle. First, we need to find the angular frequency (ω). We can do this using the formula: ω = sqrt(k/m) where k is the spring constant (16 N/m) and m is the mass (2 kg). ω = sqrt(16/2) = sqrt(8) = 2 * sqrt(2) rad/s Now we need to find the amplitude (A) and phase angle (φ). We know the initial displacement (x0) is 0.3 meters and the initial velocity (v0) is 0.2 meters/second. We can use these values to find A and φ: x0 = A * cos(φ) v0 = -A * ω * sin(φ) Solving these equations simultaneously, we get: A = sqrt(x0^2 + (v0/ω)^2) φ = arctan(-v0/(ω * x0)) A = sqrt(0.3^2 + (0.2/(2 * sqrt(2)))^2) = 0.3 m φ = arctan(-0.2/(2 * sqrt(2) * 0.3)) ≈ -0.225 radians Now we can write the equation of motion: x(t) = 0.3 * cos(2 * sqrt(2) * t - 0.225) To find the frequency (f) and period (T) of the oscillations, we can use the following formulas: f = ω / (2 * π) T = 1 / f f = (2 * sqrt(2)) / (2 * π) ≈ 0.225 Hz T = 1 / 0.225 ≈ 4.44 seconds So, the equation of motion for the mass is x(t) = 0.3 * cos(2 * sqrt(2) * t - 0.225), and the frequency and period of its oscillations are approximately 0.225 Hz and 4.44 seconds, respectively.,To determine the equation of motion for the given mechanical system, we can use Hooke's Law and Newton's second law of motion. Hooke's Law states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position (F = -kx), where F is the force, k is the spring constant, and x is the displacement. Newton's second law of motion states that the force acting on an object is equal to the mass of the object times its acceleration (F = ma), where a is the acceleration. Since the force acting on the mass is provided by the spring, we can write the equation as: ma = -kx Now, we can rearrange this to get the equation of motion: a = -(k/(m)) * x We are given the spring constant k = 16 N/m and the mass m = 2 kg. Plugging these values into the equation: a = -(16/(2)) * x a = -8x The equation of motion for the mass is: x''(t) = -8x(t) This is a simple harmonic oscillator equation, which can be solved using the following form: x(t) = A * cos(ωt + φ) Where A is the amplitude, ω is the angular frequency, and φ is the phase angle. Using the initial conditions, we know the mass is initially at a displacement of x(0) = 0.3 m and has a velocity of x'(0) = 0.2 m/s. Plugging these values into the equation: 0.3 = A * cos(φ) and 0.2 = -8 * A * sin(φ) Since the mass is at its maximum displacement, the cosine term is equal to 1, and we can find the amplitude A: A = 0.3 Now, we can use the second initial condition to find the phase angle φ: 0.2 = -8 * 0.3 * sin(φ) sin(φ) = 1/24 The phase angle is: φ = arcsin(1/24) Now, we can find the angular

answer:To find the displacement of the mass as a function of time, we can use the equation for simple harmonic motion: x(t) = A * cos(ωt + φ) where x(t) is the displacement at time t, A is the amplitude, ω is the angular frequency, and φ is the phase angle. First, let's find the angular frequency (ω) using the given spring constant (k) and mass (m): ω = √(k/m) ω = √(12 N/m / 0.1 kg) = √120 s^(-1) Next, we need to find the amplitude (A) and phase angle (φ). Since the mass is initially at rest at a displacement of 0.05 meters from the equilibrium position, the amplitude is 0.05 meters. The mass starts to move in the opposite direction of the initial displacement, so the phase angle is π radians (the angle between the initial position and the first peak). Now we can write the expression for the displacement of the mass as a function of time: x(t) = 0.05 * cos(√120 * t + π) To calculate the period of the motion, we can use the formula: T = 2π / ω T = 2π / √120 s^(-1) ≈ 0.5502 s So, the displacement of the mass as a function of time is: x(t) = 0.05 * cos(√120 * t + π) And the period of the motion is approximately: T ≈ 0.5502 s,For a mass-spring system experiencing simple harmonic motion, the displacement of the mass as a function of time can be described by the equation: x(t) = A * cos(ωt + φ) where x(t) is the displacement at time t, A is the amplitude of the motion, ω is the angular frequency, and φ is the phase angle. In this case, the mass is initially at rest at a displacement of 0.05 meters from the equilibrium position, so the amplitude A = 0.05 meters. Since the mass is initially at its maximum displacement, the phase angle φ = 0. The angular frequency ω can be found using the formula: ω = sqrt(k/m) where k is the spring constant and m is the mass. Plugging in the given values: ω = sqrt(12 N/m / 0.1 kg) = sqrt(120 s^(-2)) = 10.95 rad/s Now we can write the expression for the displacement as a function of time: x(t) = 0.05 * cos(10.95t) To find the period T of the motion, we can use the relationship between angular frequency and period: ω = 2π/T Solving for T: T = 2π/ω = 2π/10.95 ≈ 0.573 s So the period of the motion is approximately 0.573 seconds., For simple harmonic motion, the displacement of the mass as a function of time can be described using the following equation: x(t) = A * cos(ωt + φ) where A is the amplitude, ω is the angular frequency, and φ is the phase shift. The amplitude is given by the initial displacement, so A = 0.05 m. The angular frequency can be found using the formula ω = sqrt(k/m), where k is the spring constant and m is the mass: ω = sqrt(12 N/m / 0.1 kg) ω = sqrt(120 rad^2/s^2) ω = 10.9545 rad/s Since the mass is released from rest, there is no phase shift (φ = 0). So the displacement of the mass as a function of time is: x(t) = 0.05 m * cos(10.9545 rad/s * t) The period of the motion can be found using the formula T = 2π/ω: T = 2π rad / 10.9545 rad/s T ≈ 0.584 s