Question

A spring is stretched \(10 \mathrm{cm}\) by a force of 3 newtons. A mass of \(2 \mathrm{kg}\) 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 \(5 \mathrm{m} / \mathrm{sec}\). If the mass is pulled down \(5 \mathrm{cm}\) below its equilibrium position and given an initial downward velocity of \(10 \mathrm{cm} / \mathrm{sec},\) determine its position \(u\) at any time \(t\) Find the quasi frequency \(\mu\) and the ratio of \(\mu\) to the natural frequency of the corresponding undamped motion.

Short answer

Question: Calculate the position of the mass at any time, the quasi frequency, and the ratio of the quasi frequency to the natural frequency for the given spring and damper system. Answer: The position of the mass at any time is given by: $$ u(t) = -\frac{1}{90}e^{-0.3t}\left(89e^{3t}+e^{-3t}\right). $$ The quasi frequency is \(\mu =\sqrt{60}\,\mathrm{rad/s}\). The ratio of the quasi frequency to the natural frequency is 2.

Step-by-step solution

Write down Hooke's Law and the damping force equation

For a spring, Hooke's Law can be written as: $$ F_{s} = -kx, $$ where: \(F_{s}\) is the force exerted by the spring, \(k\) is the spring constant, \(x\) is the displacement from the equilibrium position. For a viscous damper, the damping force equation is: $$ F_{d} = -cv, $$ where: \(F_{d}\) is the force exerted by the damper, \(c\) is the damping coefficient, \(v\) is the velocity of the mass.

Calculate the spring constant\(k\).

Using Hooke's Law with the given values: $$ 3\,\mathrm{N}\,=-k(0.1\,\mathrm{m}) $$ Solve for k: $$ k = -30\,\mathrm{N/m} $$

Calculate the damping coefficient \(c\).

Using the damping force equation with the given values: $$ 3\,\mathrm{N}\, = -c(5\,\mathrm{m/s}) $$ Solve for c: $$ c = -0.6\,\mathrm{Ns/m} $$

Write the equation of motion.

The equation of motion is the sum of the forces acting on the mass: $$ m\frac{d^2u}{dt^2} = F_{s} + F_{d} $$ Using the derived expressions for \(F_s\) and \(F_d\), we get: $$ 2\frac{d^2u}{dt^2} = -30u - 0.6\frac{du}{dt} $$

Solve the differential equation for \(u(t)\) and apply initial conditions.

Solve the second-order linear ordinary differential equation (ODE) for u(t) and apply the given initial conditions: $$ u(0) = -0.05\,\text{m}\, \quad \text{and} \quad\frac{du}{dt}(0) = -0.10\,\mathrm{m/s} $$ The general solution of the ODE is written as: $$ u(t) = A e^{-\lambda_1 t} + B e^{-\lambda_2 t}, $$ where \(A, B, \lambda_1,\) and \(\lambda_2\) are constants to be determined. Applying the initial conditions and solving for the constants, we find the specific solution: $$ u(t) = -\frac{1}{90}e^{-0.3t}\left(89e^{3t}+e^{-3t}\right) $$

Calculate the quasi frequency and the ratio to the natural frequency.

The quasi frequency (μ) can be found as the imaginary part of the complex eigenvalue of the ODE: $$ \mu = \sqrt{4k/m - {\left(c/m\right)} ^2}\,=\, \sqrt{4(30)/2 - {\left(-0.6/2\right)}^2}\,=\, \sqrt{60}\,\mathrm{rad/s} $$ The natural frequency of the corresponding undamped motion is: $$ \omega_n = \sqrt{k/m} = \sqrt{30/2}\,=\, \sqrt{15}\,\mathrm{rad/s} $$ The ratio of the quasi frequency to the natural frequency is: $$ \frac{\mu}{\omega_n} = \frac{\sqrt{60}}{\sqrt{15}} = \sqrt{\frac{60}{15}} = 2 $$ Thus, the mass position u(t) is given by the equation: $$ u(t) = -\frac{1}{90}e^{-0.3t}\left(89e^{3t}+e^{-3t}\right), $$ the quasi frequency is \(\mu =\sqrt{60}\,\mathrm{rad/s}\), and the ratio of the quasi frequency to the natural frequency is 2.

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes the behavior of springs. It states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, it is expressed as \( F_s = -kx \), where \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement.

The negative sign indicates that the force is restorative, meaning it acts in the opposite direction of the displacement. This causes the mass attached to the spring to move back towards its equilibrium position.

• Spring constant \( k \) determines the stiffness of the spring. A larger \( k \) value means the spring is stiffer.
• Displacement \( x \) is measured from the equilibrium position, meaning it's the position of the mass when no external force is acting on the spring.

Understanding Hooke's Law is crucial for solving problems involving springs and oscillatory systems.

Damping Force

A damping force is a resistance force that acts on an oscillating object, reducing its motion over time. Unlike the spring force, which is proportional to displacement, the damping force is proportional to velocity. It is mathematically described as \( F_d = -cv \), where \( F_d \) is the damping force, \( c \) is the damping coefficient, and \( v \) is the velocity.

The negative sign indicates that damping force opposes motion, working to reduce the system's energy.

• Damping coefficient \( c \) describes how much damping is present in the system. A larger \( c \) implies more resistance, causing the motion to diminish faster.
• Velocity \( v \) is the speed at which the mass is moving, and it directly impacts the amount of damping force applied.

Damping forces are essential for understanding how real-world systems return to stability, especially in mechanical and civil engineering.

Quasi Frequency

The quasi frequency is associated with damped oscillatory systems. It reflects the modified frequency of a system when damping is present. It's determined by finding the imaginary part of the complex eigenvalue from the differential equation describing the motion.

• In a damped system, the quasi frequency is derived as \( \mu = \sqrt{\frac{4k}{m} - \left(\frac{c}{m}\right)^2} \).
• This represents how the system would oscillate if damping were reduced to zero.

Quasi frequency is important because it helps describe how the system behaves under damping. In practical applications, it considers how systems might vibrate differently due to damping present. It essentially tells how fast the vibration occurs in the presence of damping.

Natural Frequency

The natural frequency of an undamped system is its inherent oscillation rate when not subjected to any external force or damping. It indicates how a system would vibrate if left to its own devices without any resistance. This frequency serves as a reference for comparison with the system's behavior once damping or other forces are introduced.

Natural frequency is calculated using the formula \( \omega_n = \sqrt{\frac{k}{m}} \), where \( \omega_n \) is the natural frequency, \( k \) is the spring constant, and \( m \) is the mass.

• This frequency sets the baseline for understanding how additional forces change the system's behavior.
• The natural frequency is crucial in design and analysis, allowing engineers to predict the system's response under different circumstances.

Understanding natural frequency helps in designing systems like bridges, buildings, and machinery to avoid resonance, which could lead to failure.