Question

A unit mass hangs in equilibrium from a spring with constant \(k=1 / 16\). Starting at \(t=0\), a force \(F(t)=3 \sin t\) is applied to the mass. Find its displacement for \(t>0 .\)

Short answer

The general solution for the displacement of the mass in the given spring-mass system with an external force, for \(t > 0\) is: $$x(t) = \frac{1}{15}\cos{t} - \frac{48}{15}\sin{t} + C_1\cos{\frac{t}{4}} + C_2\sin{\frac{t}{4}}$$ where \(C_1\) and \(C_2\) are constants that depend on initial conditions.

Step-by-step solution

Setting up the equation of motion

The equation of motion for a forced spring-mass system is given by: $$m \frac{d^2 x}{dt^2} + kx = F(t)$$ We are given that the mass is \(m=1\) unit, the spring constant is \(k=\frac{1}{16}\), and the external force is \(F(t)=3\sin{t}\). Substitute these values into the equation of motion: $$\frac{d^2 x}{dt^2} + \frac{1}{16}x = 3 \sin{t}$$

Solving the differential equation

We have a second-order non-homogeneous linear differential equation. To solve this, we will find the particular solution first and calculate the complementary solution after. To find the particular solution, we assume a solution in the form: $$x_p(t) = A\cos{t} + B\sin{t}$$ where A and B are constants to be found. Calculate the first and second derivatives: $$\frac{dx_p(t)}{dt} = -A\sin{t} + B\cos{t}$$ $$\frac{d^2 x_p(t)}{dt^2} = -A\cos{t} - B\sin{t}$$ Now, substitute \(x_p(t)\) and its second derivative into the given differential equation: $$ -A\cos{t} - B\sin{t} + \frac{1}{16}(A\cos{t} + B\sin{t}) = 3\sin{t}$$ Comparing the coefficients of \(\cos{t}\) and \(\sin{t}\), we have: $$(-1 + \frac{1}{16})A = 0 \Rightarrow A = \frac{1}{15}$$ $$(-1 + \frac{1}{16})B = 3 \Rightarrow B = -\frac{48}{15}$$ Thus, the particular solution is: $$x_p(t) = \frac{1}{15}\cos{t} - \frac{48}{15}\sin{t}$$ Now, let's find the complementary solution. For this, we consider the related homogeneous equation: $$\frac{d^2 x}{dt^2} + \frac{1}{16}x = 0$$ This equation has the general solution: $$x_c(t) = C_1\cos{\omega t} + C_2\sin{\omega t}$$ where \(C_1\) and \(C_2\) are constants, and \(\omega = \sqrt{\frac{1}{16}} = \frac{1}{4}\). As there are no boundary conditions specified, we cannot find the values of \(C_1\) and \(C_2\).

Constructing the general solution

Combine the particular and complementary solutions to form the general solution for the given problem: $$x(t) = x_p(t) + x_c(t) = \frac{1}{15}\cos{t} - \frac{48}{15}\sin{t} + C_1\cos{\frac{t}{4}} + C_2\sin{\frac{t}{4}}$$ This is the displacement of the mass for \(t > 0\). Without further information, we cannot determine the specific values for \(C_1\) and \(C_2\).

Key concepts

Spring-Mass System

The spring-mass system is a classic physics model used to describe a mass attached to a spring. This setup can vibrate when disturbed. Imagine a block hanging from a spring. When you pull it down or push it up, it moves up and down repetitively. This motion is governed by forces acting on the mass.

• The spring itself tries to return the mass to its equilibrium position.
• Inertia causes the mass to overshoot its resting point and keep moving.

In our problem, the system involves a unit mass attaching to a spring with a spring constant, denoted by \( k = \frac{1}{16} \). The spring constant defines how stiff the spring is. A smaller value implies a softer spring, meaning it stretches or compresses more easily. By understanding the spring-mass system, we can set up equations that forecast how the mass will respond over time.

Forced Vibration

Forced vibration refers to the motion that occurs when an external force keeps acting on the spring-mass system. Unlike free vibration, where movement happens due to the system's initial disturbance and then subsides, forced vibration requires a continuous input of energy to the system.
In our example, the external force applied is a sinusoidal force \( F(t) = 3 \sin t \). This means that the force changes over time, following a wave-like pattern. It does not allow the system to return to its natural equilibrium easily. Instead, the system oscillates at the frequency of the applied force.
The inclusion of this force changes the dynamics of the system. To predict the system's displacement, we need to incorporate this force into our equations. This complexity makes finding solutions for such systems an insightful exercise for students.

Particular Solution

Finding a particular solution is a key step to solve non-homogeneous differential equations. It's where we determine a specific solution that fits the non-homogeneous nature of the differential equation, meaning it takes the external force into account.
In our problem, we assumed a solution of the form \( x_p(t) = A\cos{t} + B\sin{t} \). By substituting this guess into the differential equation, we solve for the coefficients \( A \) and \( B \).

• By equating coefficients, we find \( A = \frac{1}{15} \).
• For \( B \), solving gives us \( B = -\frac{48}{15} \).

This particular solution will eventually help describe how the mass moves due to the external force over time. The particular solution is crucial in crafting the final displacement equation.

Complementary Solution

The complementary solution provides the part of the system's response that stems from the system's own properties. It's related to the homogeneous version of the differential equation, where no external force is acting on the system.
For our case, the associated homogeneous equation is \( \frac{d^2 x}{dt^2} + \frac{1}{16}x = 0 \). Solving this equation leads to solutions involving trigonometric functions:\[ x_c(t) = C_1\cos{\frac{t}{4}} + C_2\sin{\frac{t}{4}} \]
Here, \( C_1 \) and \( C_2 \) are constants that define the exact nature of this part of the response. They depend on initial conditions that weren't specified in our problem. The complementary solution captures the system's natural oscillation tendencies independent of the external force. When combined with the particular solution, it forms the complete picture of how the system behaves.