Problem 80 - College Physics

Question

A 2.0-kg mass resting on a horizontal frictionless surface is connected to a fixed spring. The mass is displaced \(16 \mathrm{~cm}\) from its equilibrium position and released. At \(t=0.50 \mathrm{~s},\) the mass is \(8.0 \mathrm{~cm}\) from its equilibrium position (and has not passed through it yet). (a) What is the period of oscillation of the mass? (b) What are the speed of the mass and the force on the mass at \(t=0.50 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

(a) The period is 3.0 s. (b) Speed is 0.289 m/s; force is -0.28 N.

Step 1: Understand the Problem

You are given a mass-spring system, where a 2.0 kg mass is attached to a spring on a frictionless surface. The mass is initially displaced 16 cm from the equilibrium position. We need to find the period of oscillation, speed, and force on the mass at \(t = 0.50 \, \mathrm{s}\).

Step 2: Determine the Amplitude

The initial displacement of the mass is the maximum displacement, or the amplitude, \(A\). Given that the mass is displaced by 16 cm, we have \(A = 0.16 \, \mathrm{m}\).

Step 3: Use the Equation for Simple Harmonic Motion

In simple harmonic motion, displacement \(x\) as a function of time \(t\) is given by \(x(t) = A \cdot \cos(\omega t + \phi)\), where \(\omega\) is the angular frequency and \(\phi\) is the phase constant. At \(t=0\), \(x = 0.16 \, \mathrm{m}\) so \(\phi = 0\).

Step 4: Calculate Angular Frequency

We know the displacement at \(t = 0.50 \, \mathrm{s}\) is \(x = 0.08 \, \mathrm{m}\). Substitute this into the harmonic motion equation: \(0.08 = 0.16 \cdot \cos(\omega \cdot 0.50)\). Solving for \(\omega\), we find: \(\omega \cdot 0.50 = \cos^{-1}(0.5) = \frac{\pi}{3}\), thus \(\omega = \frac{2\pi}{3} \, \mathrm{rad/s}\).

Step 5: Calculate Period of Oscillation

The period \(T\) of oscillation is related to angular frequency by \(T = \frac{2\pi}{\omega}\). Substituting \(\omega = \frac{2\pi}{3}\), we find that \(T = \frac{2\pi}{\frac{2\pi}{3}} = 3.0 \, \mathrm{s}\).

Step 6: Calculate Speed of the Mass

The speed \(v\) in simple harmonic motion is given by \(v = \omega \sqrt{A^2 - x^2}\). Substituting \(\omega = \frac{2\pi}{3} \, \mathrm{rad/s}\), \(A = 0.16 \, \mathrm{m}\), and \(x = 0.08 \, \mathrm{m}\), we have \(v = \frac{2\pi}{3} \cdot \sqrt{0.16^2 - 0.08^2} \, \mathrm{m/s} = \frac{2\pi}{3} \cdot 0.138 \approx 0.289 \, \mathrm{m/s}\).

Step 7: Calculate the Force on the Mass

The force \(F\) on the mass is given by Hooke's Law, \(F = -kx\). The spring constant \(k\) can be found using \(\omega = \sqrt{\frac{k}{m}}\). Substituting \(\omega = \frac{2\pi}{3}\) and \(m = 2.0 \, \mathrm{kg}\), solve for \(k\): \(k = \left(\frac{2\pi}{3}\right)^2 \cdot 2.0 = \frac{8\pi^2}{9}\). Hence, \(F = -\left(\frac{8\pi^2}{9}\right) \cdot 0.08 = -0.28 \, \mathrm{N}\).

Key Concepts

Mass-Spring System

A mass-spring system is an essential concept in understanding simple harmonic motion. Imagine a weight, or mass, attached to a spring on a surface where there is no friction. This system is often used to model how objects move back and forth in a regular pattern, called oscillation.

The mass in our example is 2.0 kg, and it moves on a horizontal surface. When the spring is either compressed or stretched from its resting, or equilibrium, position, it exerts a force opposite to the direction of displacement. This force tries to return the mass to its original position, creating oscillations.

The initial displacement of 16 cm in this system, when the mass is pulled and released, directly affects how the system moves. This is because the maximum distance from the equilibrium position determines the amplitude of oscillation.

Oscillation Period

The oscillation period is the time it takes for the mass to complete one full cycle of its motion after being displaced. In simple harmonic motion, this period is constant and does not change with time.

In our setup, the period can be found using the formula for angular frequency, since period \(T = \frac{2\pi}{\omega}\).

Here, angular frequency \(\omega\) was deduced to be \(\frac{2\pi}{3}\) rad/s from the step-by-step solution. Thus, substituting into the formula gives the period \(T = 3.0 \, \text{s}\). This value means that every 3 seconds, the mass completes a cycle of moving from one side of the quotient path to the other and back again.

Hooke's Law

Hooke's Law is a principle that describes the force exerted by a spring when it is displaced from its equilibrium. According to it, the force \(F\) is directly proportional to the displacement \(x\) but acts in the opposite direction. This is expressed as \(F = -kx\), where \(k\) is the spring constant.

In our exercise, we've determined the spring constant using the angular frequency. With \(\omega = \frac{2\pi}{3}\) and knowing \(m\), we found \(k = \frac{8\pi^2}{9}\).

When the displacement of 0.08 m at \(t = 0.50 \, \text{s}\) is plugged into Hooke's Law, the resultant force on the mass is \(F = -0.28 \, \text{N}\). This force moves the mass back toward its starting point, continuing the oscillation until another external force interrupts the motion.

Angular Frequency

Angular frequency, denoted by \(\omega\), measures how quickly an object travels through its cycle in simple harmonic motion. It's tied to the oscillation properties of the system, determining how many cycles are completed in a given period.

In our context, angular frequency is crucial for finding both the period of oscillation and other dynamic properties like velocity and force. By knowing the initial displacement and the position at a specific time, we solved for \(\omega = \frac{2\pi}{3}\) rad/s.

This value not only allowed us to calculate our oscillation period but also facilitated the determination of the mass's velocity at a particular point using \(v = \omega \sqrt{A^2 - x^2}\). Angular frequency's value, balanced with mass and spring constant, intricately connects to describe the mass-spring system's whole motion.