Oscillations: Particle On A Spring Under Force

Hey physics fanatics! Ever wondered what happens when you mess with a classic setup, like a particle on a spring? Well, today we're diving deep into a super interesting scenario: a particle of mass m chilling at rest at the end of a spring with a force constant k, hanging from a fixed support. You know, the usual setup. But then, things get exciting! At t = 0, we throw a wrench in the works by applying a constant downward force F. This force isn't just a fleeting guest; it sticks around for a specific duration, t₀. After this time, the force is removed. Our mission, should we choose to accept it (and we totally do!), is to show that, after the force is removed, the displacement of the particle can be described by a particular equation. This isn't just about crunching numbers, guys; it's about understanding the dynamics of systems, how they respond to external influences, and how they settle back into their natural rhythm. We'll break down the physics step-by-step, making sure we grasp every concept along the way. Get ready to explore simple harmonic motion, damped oscillations, and forced oscillations – all bundled into one awesome problem!

Setting the Stage: The Undisturbed System

Before we even think about applying that external force F, let's get a handle on our system in its natural state. We have a particle of mass m attached to a spring with a force constant k. This spring is hanging vertically from a fixed support. When the system is just hanging there, at rest, the spring is stretched by some amount. Let's call this equilibrium position $y=0$. At this equilibrium, the upward force exerted by the spring is exactly balancing the downward force of gravity on the mass. So, $ky_{eq} = mg$, where $y_{eq}$ is the equilibrium stretch. This is crucial because whenever we talk about displacement from this point onwards, we're measuring it from this equilibrium position. It simplifies our equations immensely. Now, if we were to pull this mass down a bit and let it go, it would oscillate up and down with a specific frequency, $\omega = \sqrt{k/m}$. This natural frequency dictates how fast the system wants to oscillate when left to its own devices. It's the inherent rhythm of our mass-spring system. Understanding this baseline behavior is like learning the alphabet before you can write a novel; it's fundamental to grasping how the system will react when we introduce external forces. So, before $t=0$, our particle is just hanging out, perfectly still, at its equilibrium position. No motion, no drama, just gravity and spring force in perfect harmony. This initial condition is our starting point, our blank canvas upon which we'll paint the complex dance of motion caused by the applied force.

The Perturbation: Applying the Force F

Alright, so things are calm, but not for long! At $t=0$, we introduce our external agent: a constant downward force F. This force is applied to the particle of mass m and, crucially, it acts for a finite time t₀. This means our system, which was happily at rest, is now being forced to move. The equation of motion for the particle, while this force is active (i.e., for $0 \le t \le t₀$), is modified. Remember Newton's second law: $F_{net} = ma$. In our case, the net force is the sum of the gravitational force ($mg$), the spring force ($-ky$, where $y$ is the displacement from the unstretched position, or more conveniently, $-k(y_{initial} + y_{new})$ if $y$ is from equilibrium and $y_{initial}$ is the equilibrium stretch), and the applied force ($F$). If we define $y$ as the displacement from the equilibrium position (where $ky_{eq} = mg$), then the gravitational force and the spring force at equilibrium cancel out. The equation of motion becomes: $m\ddot{y} = F - ky$. Rearranging this, we get $m\ddot{y} + ky = F$. This is the differential equation governing the motion of the particle while the force F is applied. Notice how the applied force F acts as a constant driving term. This term shifts the equilibrium position and introduces a new type of motion. The system will not only oscillate but also exhibit a steady-state displacement due to this constant force. We need to solve this second-order non-homogeneous differential equation. The general solution will consist of a complementary solution (which describes the natural oscillations) and a particular solution (which describes the steady-state response to the constant force F). Solving this requires us to consider initial conditions at $t=0$. Since the particle starts at rest at the equilibrium position, our initial conditions are $y(0) = 0$ and $\dot{y}(0) = 0$. These initial conditions will allow us to determine the constants of integration that arise from solving the differential equation. The duration for which this force acts, $t₀$, is also a critical parameter. The state of the system (its position and velocity) at time $t₀$ will become the initial conditions for the next phase of motion, where the applied force is no longer present.

The Transition: Force Removed at t₀

Now, the exciting part! At time $t = t₀$, the constant downward force F is removed. This is a critical juncture in our problem. What happens to the particle's motion after this point? The differential equation governing the motion changes. For $t > t₀$, the applied force F is no longer acting on the mass. So, the equation of motion reverts back to the standard form for a simple harmonic oscillator, but with a crucial difference: the initial conditions for this new phase of motion are not the ones we started with at $t=0$. Instead, the initial conditions for $t > t₀$ are the position and velocity of the particle at the exact moment the force was removed, i.e., at $t = t₀$. Let $y(t₀)$ and $\dot{y}(t₀)$ be the displacement and velocity of the particle at time $t₀$. These values are determined by the solution of the equation of motion during the interval $0 \le t \le t₀$. So, the first step is to solve $m\ddot{y} + ky = F$ with initial conditions $y(0) = 0$ and $\dot{y}(0) = 0$. Once we have the solution $y(t)$ for $0 \le t \le t₀$, we can evaluate $y(t₀)$ and $\dot{y}(t₀)$. These values then become the initial conditions for the subsequent motion described by the equation $m\ddot{y} + ky = 0$ for $t > t₀$. This equation represents free oscillations. The general solution for this homogeneous equation is of the form $y(t) = A \cos(\omega t) + B \sin(\omega t)$, where $\omega = \sqrt{k/m}$. The constants A and B are determined by the initial conditions $y(t₀)$ and $\dot{y}(t₀)$. This transition is key. The motion after $t₀$ is a superposition of the system's natural tendency to oscillate at its own frequency $\omega$ and the imprint of the previous forced motion. The amplitude and phase of these free oscillations are directly dictated by the state of the system when the external force vanished. It's like giving a swing a push for a few seconds; the swing will continue to move with its own natural period, but the height it reaches and its subsequent motion depend entirely on how hard and for how long you pushed it. The removal of the force doesn't instantly reset the system; it carries the memory of the applied force in its momentum and position.

Deriving the Displacement After Force Removal

Alright, guys, let's get down to the nitty-gritty and actually show the displacement after the force is removed. We need to solve the problem in two stages.

Stage 1: $0 \le t \le t₀$ (Force F is applied)

The equation of motion is $m\ddot{y} + ky = F$. The general solution is the sum of the complementary solution $y_c(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t)$ (where $\omega = \sqrt{k/m}$) and a particular solution $y_p(t)$. For a constant force F, a good guess for the particular solution is a constant, say $y_p = A$. Substituting into the equation: $m(0) + kA = F$, so $A = F/k$. This is the new equilibrium position under the influence of force F.

The total solution is $y(t) = C_1 \cos(\omega t) + C_2 \sin(\omega t) + F/k$.

Now, we apply the initial conditions at $t=0$: $y(0) = 0$ and $\dot{y}(0) = 0$.

From $y(0)=0$: $C_1 \cos(0) + C_2 \sin(0) + F/k = 0 implies C_1 + F/k = 0

So, $C_1 = -F/k$.

To find $C_2$, we need the velocity $\dot{y}(t)$: $\dot{y}(t) = -C_1 \omega \sin(\omega t) + C_2 \omega \cos(\omega t)$.

From $\dot{y}(0)=0$: $-C_1 \omega \sin(0) + C_2 \omega \cos(0) = 0

This gives $C_2 \omega = 0$. Since $\omega \ne 0$, we must have $C_2 = 0$.

So, the displacement during the time the force is applied is: $y(t) = \frac{F}{k} (1 - \cos(\omega t))$ for $0 \le t \le t₀$.

This equation tells us how the particle moves until time $t₀$. It starts from rest and moves to a maximum displacement.

Now, we need the position and velocity at $t = t₀$ to serve as initial conditions for the next stage.

$y(t₀) = \frac{F}{k} (1 - \cos(\omega t₀))$

$\dot{y}(t) = \frac{F}{k} (\omega \sin(\omega t))$

$\dot{y}(t₀) = \frac{F}{k} \omega \sin(\omega t₀)$

Stage 2: $t > t₀$ (Force F is removed)

The equation of motion is now $m\ddot{y} + ky = 0$. The general solution is $y(t) = D_1 \cos(\omega t) + D_2 \sin(\omega t)$.

We use the values from Stage 1 at $t = t₀$ as our initial conditions for this equation.

Let's shift our time variable for convenience. Let $t' = t - t₀$. So, when $t=t₀$, $t'=0$. The equation becomes $y(t') = D_1 \cos(\omega (t'+t₀)) + D_2 \sin(\omega (t'+t₀))$.

Using the standard form $y(t') = A' \cos(\omega t') + B' \sin(\omega t')$ and applying initial conditions at $t'=0$ (which corresponds to $t=t₀$):

$y(0) = y(t₀) = \frac{F}{k} (1 - \cos(\omega t₀)) = A' \cos(0) + B' \sin(0)

So, $A' = \frac{F}{k} (1 - \cos(\omega t₀))$.

Now for the velocity. The velocity in terms of $t'$ is $\dot{y}(t') = -A' \omega \sin(\omega t') + B' \omega \cos(\omega t')$.

At $t'=0$ (which is $t=t₀$):

$\dot{y}(0) = \dot{y}(t₀) = \frac{F\omega}{k} \sin(\omega t₀) = -A' \omega \sin(0) + B' \omega \cos(0)

So, $\dot{y}(t₀) = B' \omega$.

$B' = \frac{\dot{y}(t₀)}{\omega} = \frac{1}{\omega} \frac{F\omega}{k} \sin(\omega t₀) = \frac{F}{k} \sin(\omega t₀)$.

Substituting $A'$ and $B'$ back into $y(t')$:

$y(t') = \frac{F}{k} (1 - \cos(\omega t₀)) \cos(\omega t') + \frac{F}{k} \sin(\omega t₀) \sin(\omega t')$

$y(t') = \frac{F}{k} [\cos(\omega t') - \cos(\omega t₀) \cos(\omega t') + \sin(\omega t₀) \sin(\omega t')]$

Using the trigonometric identity $\cos(A-B) = \cos A \cos B + \sin A \sin B$, we can rearrange the terms inside the bracket. Let $A = \omega t'$ and $B = \omega t₀$. Then $\cos(\omega t') \cos(\omega t₀) - \sin(\omega t') \sin(\omega t₀) = \cos(\omega t' + \omega t₀)$.

Wait, that's not quite right. Let's re-examine the identity. We have $- \cos(\omega t₀) \cos(\omega t') + \sin(\omega t₀) \sin(\omega t')$.

Let's use $\cos(A+B) = \cos A \cos B - \sin A \sin B$. Not quite.

Let's use $\cos(A-B) = \cos A \cos B + \sin A \sin B$. Not quite.

Consider the identity $\cos(X - Y) = \cos X \cos Y + \sin X \sin Y$. Let $X = \omega t'$ and $Y = \omega t₀$. Then $\cos(\omega t' - \omega t₀) = \cos(\omega t') \cos(\omega t₀) + \sin(\omega t') \sin(\omega t₀)$.

We have: $y(t') = \frac{F}{k} [\cos(\omega t') - (\cos(\omega t₀) \cos(\omega t') - \sin(\omega t₀) \sin(\omega t'))]$

$y(t') = \frac{F}{k} [\cos(\omega t') - \cos(\omega t' + \omega t₀)]$

Substituting back $t' = t - t₀$:

$y(t - t₀) = \frac{F}{k} [\cos(\omega (t - t₀)) - \cos(\omega t)]$

This is the displacement of the particle after the force is removed, expressed in terms of the time elapsed since the force was removed ($t-t₀$) and the total time elapsed since the beginning ($t$). This equation is valid for $t > t₀$.

Let's check this. At $t=t₀$ (so $t'=0$), $y(0) = \frac{F}{k} [\cos(0) - \cos(\omega t₀)] = \frac{F}{k} (1 - \cos(\omega t₀))$, which matches $y(t₀)$ from Stage 1. Good!

What about velocity at $t=t₀$? Differentiating $y(t - t₀)$ with respect to $t$:

$\frac{dy}{dt} = \frac{F}{k} [-\omega \sin(\omega (t - t₀)) - (-\omega \sin(\omega t))] = \frac{F\omega}{k} [\sin(\omega t) - \sin(\omega (t - t₀))]$

At $t=t₀$: $\frac{dy}{dt}|_{t=t₀} = \frac{F\omega}{k} [\sin(\omega t₀) - \sin(0)] = \frac{F\omega}{k} \sin(\omega t₀)$, which matches $\dot{y}(t₀)$ from Stage 1. Excellent!

So, the final expression for the displacement after the force is removed is:

$$y(t) = \frac{F}{k} \left[ \cos(\omega (t - t₀)) - \cos(\omega t) \right] \quad \text{for } t > t₀$$

where $\omega = \sqrt{k/m}$. This formula beautifully encapsulates the subsequent oscillatory motion of the particle, driven by its initial conditions set during the force application period.

Understanding the Result

So, what does this final equation, $y(t) = \frac{F}{k} [\cos(\omega (t - t₀)) - \cos(\omega t)]$ for $t > t₀$, actually tell us? It's a bit of a mind-bender at first, with those two cosine terms, but it elegantly describes the free oscillation of the particle after the external force F has been removed. Remember, this equation is valid after time $t₀$. The term $\omega = \sqrt{k/m}$ is our familiar angular frequency, the natural 'beat' of the spring-mass system. The equation essentially shows that the motion for $t > t₀$ is a superposition of two oscillations. One oscillation seems to be phase-shifted by $t₀$, represented by $\cos(\omega (t - t₀))$, and the other is a standard cosine wave $\cos(\omega t)$.

Let's think about the amplitude. The term $F/k$ is the static displacement if the force F were applied permanently. So, the amplitude of the oscillations is somehow related to this static deflection. We can use a trigonometric identity to rewrite this expression, which might give us more physical insight. Recall the identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$.

Let $A = \omega t$ and $B = \omega (t - t₀)$. Then $\frac{A+B}{2} = \frac{\omega t + \omega t - \omega t₀}{2} = \frac{2\omega t - \omega t₀}{2} = \omega t - \frac{\omega t₀}{2}$. And $\frac{A-B}{2} = \frac{\omega t - (\omega t - \omega t₀)}{2} = \frac{\omega t₀}{2}$.

So, $\cos(\omega t) - \cos(\omega (t - t₀)) = -2 \sin\left(\omega t - \frac{\omega t₀}{2}\right) \sin\left(\frac{\omega t₀}{2}\right)$.

Therefore, our displacement equation becomes:

$y(t) = \frac{F}{k} \left[ -2 \sin\left(\frac{\omega t₀}{2}\right) \sin\left(\omega t - \frac{\omega t₀}{2}\right) \right]$

$y(t) = -2 \frac{F}{k} \sin\left(\frac{\omega t₀}{2}\right) \sin\left(\omega t - \frac{\omega t₀}{2}\right)$

This form is much more revealing! It shows that the displacement is a sinusoidal wave $\sin(\omega t - \frac{\omega t₀}{2})$ whose amplitude is $2 \frac{F}{k} \left| \sin(\frac{\omega t₀}{2}) \right|$.

This amplitude depends on the duration $t₀$ for which the force was applied. It also depends on F and k, as expected.

• If $\sin(\frac{\omega t₀}{2}) = 0$: This happens when $\frac{\omega t₀}{2} = n\pi$ for integer $n$. This means $\omega t₀ = 2n\pi$. In this case, the amplitude is zero, and the particle returns to its equilibrium position and stays there after $t₀$. This is a very specific scenario where the force was applied for exactly a half-integer multiple of the period $T = 2\pi/\omega$. For example, if $t₀ = T/2$, then $\omega t₀ = (2\pi/T)(T/2) = \pi$. Then $\sin(\pi/2) = 1$. This doesn't result in zero amplitude. Let's recheck. If $\omega t₀ = 2n\pi$, then $t₀ = nT$. This means the force was applied for an integer number of full periods. In this case, the system returns to its exact starting state (zero displacement and zero velocity) if the force was harmonic. However, here the force is constant. If $\omega t₀ = 2n\pi$, then $\sin(\frac{\omega t₀}{2}) = \sin(n\pi) = 0$. So, the amplitude is indeed zero. This means the net effect of the force over these periods results in the particle being at rest at $y=0$.
• If $\sin(\frac{\omega t₀}{2})$ is maximum (i.e., $\pm 1$): This happens when $\frac{\omega t₀}{2} = (n + 1/2)\pi$. This means $\omega t₀ = (2n+1)\pi$. The force is applied for a half-integer number of periods. In this case, the amplitude of oscillation is maximal, equal to $2F/k$. The particle is set into motion with the largest possible amplitude.

This derived equation shows that the motion after the force removal is a pure harmonic oscillation. The amplitude is determined by the duration ($t₀$) and magnitude ($F$) of the applied force, as well as the system's natural properties ($m$ and $k$). It’s a beautiful demonstration of how a transient force can leave a lasting imprint on a system, setting it into a new state of motion.

Conclusion: The Enduring Oscillation

So there you have it, folks! We've successfully navigated the twists and turns of a particle on a spring subjected to a temporary force. We started with a system at rest, introduced a constant force for a specific duration ($t₀$), and then let the system run free. The result? A beautiful, persistent oscillation described by the equation $y(t) = \frac{F}{k} [\cos(\omega (t - t₀)) - \cos(\omega t)]$ for $t > t₀$. We even transformed this into a more insightful form, $y(t) = -2 \frac{F}{k} \sin(\frac{\omega t₀}{2}) \sin(\omega t - \frac{\omega t₀}{2})$, which clearly shows the oscillatory nature and an amplitude that depends critically on the duration $t₀$ of the applied force.

This problem is a fantastic illustration of several key physics principles. Firstly, it showcases Newton's second law in action, governing the motion under different forces. Secondly, it highlights the concept of superposition, where the motion after the force is removed is a consequence of the system's response during the period the force was applied. The state of the system at $t₀$ (its position and velocity) dictates the subsequent free oscillations. Thirdly, it demonstrates the principles of simple harmonic motion and how external perturbations can excite these oscillations. The fact that the amplitude of the resulting oscillation can be zero or maximal depending on $t₀$ is a particularly neat outcome, showcasing the resonant-like behavior even with a constant force applied for a duration.

Understanding these dynamics is not just an academic exercise. It has real-world implications, from designing earthquake-resistant structures (where external forces are applied and then cease) to analyzing the behavior of mechanical systems and even in understanding the vibration of musical instruments. So, the next time you see a mass on a spring, remember that even a brief nudge can set it into a long-lasting dance!

Keep exploring, keep questioning, and keep enjoying the wonderful world of physics!