Question

A mass of \(100 \mathrm{g}\) stretches a spring \(5 \mathrm{cm}\). If the mass is set in motion from its equilibrium position with a downward velocity of \(10 \mathrm{cm} / \mathrm{sec},\) and if there is no damping, determine the position \(u\) of the mass at any time \(t .\) When does the mass first return to its equilibrium position?

Short answer

Answer: The mass first returns to its equilibrium position at approximately \(t\approx 0.3362\mathrm{s}\).

Step-by-step solution

Find the spring constant k

We know the equation for Hooke's law is \(F = k \cdot x\), where F is the force acting on the mass due to the spring, k is the spring constant, and x is the displacement from the equilibrium position. The mass is given as \(100\mathrm{g}\), which is equivalent to \(m = 0.1\mathrm{kg}\). The displacement from the equilibrium position is \(5\mathrm{cm}\), which is equivalent to \(x = 0.05\mathrm{m}\). The force acting on the mass due to the spring is \(F = m \cdot g\), where \(g = 9.81\mathrm{m/s^2}\) is the acceleration due to gravity. So, \(F = 0.1\mathrm{kg} \cdot 9.81\mathrm{m/s^2} = 0.981\mathrm{N}\) Now we can find the spring constant \(k\) using the equation: \(k = \frac{F}{x} = \frac{0.981\mathrm{N}}{0.05\mathrm{m}} = 19.62\frac{\mathrm{N}}{\mathrm{m}}\)

Find the angular frequency \(\omega\)

For a mass oscillating with simple harmonic motion, the angular frequency is given by: \(\omega = \sqrt{\frac{k}{m}}\) Substitute the values of \(k\) and \(m\) we found above: \(\omega = \sqrt{\frac{19.62\frac{\mathrm{N}}{\mathrm{m}}}{0.1\mathrm{kg}}} = \sqrt{196.2\mathrm{s^{-2}}}= 14\mathrm{s^{-1}}\)

Find the amplitude A and phase angle \(\phi\)

Let the position of the mass at any time t be represented as \(u(t)=A\cos{(\omega t + \phi)}\) We're given that at \(t=0\), the mass is in motion from equilibrium position (u=0) with a downward velocity \(10\mathrm{cm/s}\) (\(0.1\mathrm{m/s}\)). So, we have to find the amplitude A and the phase angle \(\phi\) using these initial conditions: 1. At \(t = 0\), \(u = 0 = A\cos{(\omega(0) + \phi)} = A\cos{(\phi)}\) 2. \(u'(t)=-A\omega\sin{(\omega t + \phi)}\) (the derivative of \(u\) with respect to \(t\)) At \(t=0\), \(u'(0)=-0.1\mathrm{m/s}=-A\omega\sin(\phi)\) We have two equations for A and \(\phi\). By solving, we can obtain: 1. From \(u=0\), \(A\cos{(\phi)}=0\) ⟹ \(\cos{(\phi)}=0\) ⟹ \(\phi=\frac{\pi}{2}\) 2. From \(u'(0)=-0.1\mathrm{m/s}\), \(-0.1\mathrm{m/s}=-A(14\mathrm{s^{-1}})\sin(\frac{\pi}{2})\) ⟹ \(A=\frac{0.1}{14}\mathrm{m}\)

Determine the position u of the mass at any time t

We found that \(A=\frac{0.1}{14}\mathrm{m}\) and \(\phi=\frac{\pi}{2}\). By substituting these values into the equation for \(u(t)\), we can find the position u of the mass at any time t: \(u(t)=\frac{0.1}{14}\mathrm{m}\cdot\cos{(14\mathrm{s^{-1}}t + \frac{\pi}{2})}\)

Find when the mass first returns to its equilibrium position

The mass first returns to its equilibrium position when \(u(t) = 0\). So, we have to solve the following equation for t: \(0 = \frac{0.1}{14}\mathrm{m}\cdot\cos{(14\mathrm{s^{-1}}t + \frac{\pi}{2})}\) Since the cosine function equals zero when its argument is \(\frac{\pi}{2} + n\pi\), where \(n\) is an integer, we can write: \(14\mathrm{s^{-1}}t + \frac{\pi}{2}=\frac{\pi}{2} + n\pi\) Solving for t: \(t = \frac{2\pi n -\frac{\pi}{2}}{14}=\frac{\pi(4n-1)}{28}\) Since we are looking for the first return to the equilibrium position, we can set \(n=1\): \(t = \frac{\pi(4(1)-1)}{28} = \frac{3\pi}{28}\mathrm{s}\approx 0.3362\mathrm{s}\) So, the mass first returns to its equilibrium position at \(t\approx 0.3362\mathrm{s}\).

Key concepts

Hooke's Law

Hooke's Law is a fundamental principle that relates the force exerted by a spring to the displacement from its equilibrium position. Imagine a spring that you pull or compress; it either "pulls back" or "pushes back" with a force that's proportional to how far you've moved it. This law is expressed mathematically as:

• \( F = k \cdot x \)

Here, \( F\) represents the force applied to the spring, \( k \) is the spring constant, and \( x \) is the displacement from its original, relaxed position.
Understanding Hooke's Law is crucial in analyzing simple harmonic motion because it explains how springs behave when forces are applied. In the exercise, knowing the mass of 100 g and that it stretches the spring by 5 cm allows us to calculate the spring constant \( k \). This plays an essential role in determining other characteristics of the oscillation such as angular frequency and phase angle.

Spring Constant

The spring constant \( k \) tells us how stiff or flexible a spring is. It basically measures the resistance of a spring to being compressed or stretched. You can think of \( k \) as a number that indicates whether the spring is weak and easy to stretch or strong and hard to move.
In our example, we calculated \( k \) by rearranging Hooke's Law into:

• \( k = \frac{F}{x} \)

We found the force \( F \) using the formula \( F = m \cdot g \), where \( m \) is mass, and \( g \) is the acceleration due to gravity (9.81 m/s²). This resulted in \( k = 19.62 \, \mathrm{N/m} \), which tells us how much force is needed to move the spring by one meter. The spring constant is crucial for understanding how the spring helps create simple harmonic motion, as it influences the system's dynamics and characteristics.

Angular Frequency

Angular Frequency, denoted by \( \omega \), is the measure of the rate of rotation or how fast a system oscillates in simple harmonic motion. Instead of just counting how many cycles happen per second (frequency), angular frequency tells us how many radians the system moves through each second. This is given by the formula:

• \( \omega = \sqrt{\frac{k}{m}} \)

In the exercise, using the values of \( k = 19.62 \, \mathrm{N/m} \) and \( m = 0.1 \, \mathrm{kg} \), we calculate \( \omega = 14 \, \mathrm{s^{-1}} \).
This high angular frequency indicates that the mass moves quickly back and forth. Understanding \( \omega \) helps predict how the system behaves over time, especially how rapidly it returns to its starting position. It's a critical factor in characterizing any oscillating system.

Phase Angle

The phase angle \( \phi \) helps us understand the initial conditions of the oscillation. It tells where in its cycle the motion of the object begins at \( t = 0 \). Imagine starting the pendulum: sometimes you start from the middle, sometimes from one side. The phase angle captures these possibilities mathematically.
In the context of simple harmonic motion, \( \phi \) is crucial for describing the full motion using a function like \( u(t) = A \cos{(\omega t + \phi)} \). During the exercise, solving equations at \( t = 0 \) such as \( u(0) = 0 \) gave \( \phi = \frac{\pi}{2} \).
This means the motion starts from a specific point in the cycle, and \( \phi \) ensures the math accounts for that initial position, whether starting at maximum displacement, at equilibrium with a certain velocity, or anywhere in between.