Question

A body of mass \(m\) is in rectilinear motion along a horizontal axis. The resultant force acting on the body is given by \(-k x\), where \(k>0\) is a constant of proportionality and \(x\) is the distance along the axis from a fixed point \(\mathrm{O}\). The body has initial velocity \(v=v_{0}\) when \(x=x_{0}\). Apply Newton's second law in the form ( \(3.23\) ) and thus write the differential equation of motion in the form $$ m v \frac{d v}{d x}=-k x . $$ Solve the differential equation, apply the initial condition, and thus express the square of the velocity \(v\) as a function of the distance \(x\). Recalling that \(v=d x / d t\), show that the relation between \(v\) and \(x\) thus obtained is satisfied for all time \(t\) by $$ x=\sqrt{x_{0}^{2}+\frac{m v_{0}^{2}}{k}} \sin \left(\sqrt{\frac{k}{m}} t+\phi\right) $$ where \(\phi\) is a constant.

Short answer

The square of the velocity \(v\) can be expressed as a function of distance \(x\), given by: $$ v^2 = \frac{k}{m}({x_0}^2 - x^2) + {v_0}^2. $$ The relation between \(v\) and \(x\) for all time \(t\) can be described by the position function: $$ x(t) = \sqrt{{x_0}^2+\frac{m{v_0}^2}{k}}\sin(\omega t + \phi), $$ where \(\omega = \sqrt{\frac{k}{m}}\) and \(\phi\) is a constant.

Step-by-step solution

Apply Newton's second law and obtain the differential equation of motion

According to Newton's second law, the force acting on the body is equal to the product of its mass and acceleration, i.e., \(F = ma\). The given force is \(F = -kx\). Thus, we have $$ ma = -kx. $$ Recall that acceleration, \(a\), is the second derivative of position with respect to time: \(a = \frac{d^2x}{dt^2}\). Also, remember that the chain rule states that \(\frac{d^2x}{dt^2} = \frac{dx}{dt}\frac{d^2x}{dx^2} = v\frac{dv}{dx}\). Therefore, combining the given information, we have $$ m(v\frac{dv}{dx}) = -kx, $$ which is the differential equation of motion we want to solve.

Solve the differential equation

To solve the first-order differential equation \(mv\frac{dv}{dx}=-kx\), we first separate the variables as follows: $$ \frac{v\, dv}{dx}=-\frac{k}{m}x. $$ Now, integrate both sides of the equation with respect to \(x\): $$ \int v\, dv = -\int \frac{k}{m}x\, dx. $$ After integrating, we have $$ \frac{1}{2}v^2 = -\frac{k}{2m}x^2 + C, $$ where \(C\) is the constant of integration.

Apply the initial condition

We're given that the body has an initial velocity \(v_0\) when \(x = x_0\). Substituting these values into the integrated equation, we get $$ \frac{1}{2}{v_0}^2 = -\frac{k}{2m}{x_0}^2 + C. $$ Solving for \(C\), we get $$ C=\frac{1}{2}{v_0}^2+\frac{k}{2m}{x_0}^2. $$ Now, substitute the value of \(C\) back into the equation for \(v^2\): $$ v^2 = -\frac{k}{m}x^2 + \frac{1}{2}{v_0}^2+\frac{k}{2m}{x_0}^2. $$ Rearranging the equation, we express the square of the velocity \(v\) as a function of the distance \(x\): $$ v^2 = \frac{k}{m}({x_0}^2 - x^2) + {v_0}^2. $$

Find the relation between \(v\) and \(x\) for all time \(t\)

Recall the relation \(v = \frac{dx}{dt}\), which can be rewritten as \(\frac{dx}{dt} = v = \pm\sqrt{\frac{k}{m}({x_0}^2 - x^2) + {v_0}^2}\). Now, let the constant \(\omega = \sqrt{\frac{k}{m}}\). Notice that this equation can represent an oscillatory motion if we can rewrite it in the form of a sinusoidal function. By observation, we can have a suitable expression for \(x(t)\) as follows: $$ x(t) = \sqrt{{x_0}^2+\frac{m{v_0}^2}{k}}\sin(\omega t + \phi), $$ where \(\phi\) is a constant. Indeed, if we differentiate \(x(t)\) with respect to \(t\), we get $$ \frac{dx}{dt} = \omega\sqrt{{x_0}^2+\frac{m{v_0}^2}{k}}\cos(\omega t + \phi), $$ which is related to the original expression for \(v^2\). Thus, this expression for \(x(t)\) satisfies the relation between \(v\) and \(x\) for all time \(t\).

Key concepts

Newton's Second Law

Understanding Newton's Second Law is crucial for analyzing the motion of objects. It states that the force acting on an object is equal to the mass of the object times its acceleration, encapsulated in the formula: \( F = ma \). This relationship helps us determine how an object will react to a given force, such as how quickly it will speed up or slow down.
The exercise involves a body experiencing a force proportional to its distance from a fixed point: \( F = -kx \). Here, the force is directed towards the fixed point, suggesting a "restorative" force typical of harmonic motion. By setting this equal to \( ma \), we form the differential equation of the body's motion.
This foundational principle serves as the basis for more complex motion scenarios, such as the one in this exercise, where we also apply mathematical techniques to tie velocity and position into the analysis.

Rectilinear Motion

Rectilinear motion refers to motion along a straight line. The exercise focuses on an object moving along a horizontal axis. By considering such linear motion, we simplify the problem to one dimension.
In a rectilinear system, the motion of the body is described using the body's position \( x \), velocity \( v \), and acceleration \( a \), which evolve over time. In our task, the force depends on the position \(-kx\) and guides us to solve how velocity \( v \) relates to distance \( x \) but within that constant rectilinear path.
This type of motion is particularly straightforward, making it an excellent starting point when learning about forces and dynamics in physics. It allows us to isolate the effects of forces on motion without the additional complexity of curved or multi-dimensional paths.

Harmonic Motion

Harmonic motion describes an oscillatory movement, like a pendulum swinging back and forth. In our problem, the force \(-kx\) indicates a harmonic nature, as it's analogous to a spring pulling towards equilibrium.
The differential equation we derived from Newton's Second Law has solutions that naturally take the sinusoidal form, characteristic of harmonic motion. This is why the position of the body can be expressed as a sine function of time: \( x = \sqrt{x_0^2 + \frac{mv_0^2}{k}} \sin \left(\sqrt{\frac{k}{m}}t + \phi\right) \).
The term \( \sqrt{\frac{k}{m}} \) represents the angular frequency, showing how quickly the oscillations occur. Such results highlight that harmonic motion is both predictable and mathematically describable, essential in many fields from engineering to seismology.

Initial Conditions

Initial conditions provide the necessary information to uniquely solve differential equations involving motion. They specify the state of a system at the start of the observation, providing values such as initial position or velocity.
In our exercise, the initial conditions are given as \( v = v_0 \) when \( x = x_0 \). This allows us to determine the constant of integration \( C \) when solving our differential equation. Knowing the initial velocity and position helps to tailor the general solution to this specific scenario.
Initial conditions are vital in physics as they ensure that solutions to equations of motion are applicable to real-world problems. They turn the mathematically derived results into accurate descriptions of how an object will actually move under the specified forces and constraints.