Problem 88 - Calculus 3

Question

A 100-g mass stretches a spring 0.1 m. Find the equation of motion of the mass if it is released from rest from a position 20 cm below the equilibrium position. What is the frequency of this motion?

Step-by-Step Solution

Answer

Equation of motion: \( x(t) = -0.2\cos(9.9t) \); Frequency: \( 1.576 \text{ Hz} \).

Step 1: Determine the spring constant (k)

The spring constant can be found using Hooke's Law, which states that the force exerted by a spring is proportional to its extension, with the relation \( F = kx \). Given weight of the mass \( F = mg = (0.1 \text{ kg})(9.8 \text{ m/s}^2) = 0.98 \text{ N} \) and the extension \( x = 0.1 \text{ m} \), we can set \( k \cdot 0.1 = 0.98 \) which yields \( k = \frac{0.98}{0.1} = 9.8 \text{ N/m} \).

Step 2: Write the differential equation of motion

The differential equation of motion for a spring-mass system is given by \( m\frac{d^2x}{dt^2} + kx = 0 \). Substituting \( m = 0.1 \text{ kg} \) and \( k = 9.8 \text{ N/m} \), we have \( 0.1\frac{d^2x}{dt^2} + 9.8x = 0 \). Simplifying gives \( \frac{d^2x}{dt^2} + 98x = 0 \).

Step 3: Solve the differential equation for x(t)

The standard form of the solution to the differential equation \( \frac{d^2x}{dt^2} + \omega^2x = 0 \) is \( x(t) = A\cos(\omega t) + B\sin(\omega t) \), where \( \omega^2 = \frac{k}{m} = 98 \), thus \( \omega = \sqrt{98} \approx 9.9 \). Therefore, the solution is \( x(t) = A\cos(9.9t) + B\sin(9.9t) \).

Step 4: Apply initial conditions to find constants

The mass is released from rest 20 cm (0.2 m) below the equilibrium position, thus at \( t = 0 \), \( x = -0.2 \) and \( \frac{dx}{dt} = 0 \). Using these conditions, \( x(0) = A \) gives \( A = -0.2 \). The derivative \( \frac{dx}{dt} = -9.9A\sin(9.9t) + 9.9B\cos(9.9t) \) evaluated at \( t = 0 \) gives \( 0 = 9.9B \), so \( B = 0 \). Thus, \( x(t) = -0.2\cos(9.9t) \).

Step 5: Calculate the frequency of the motion

The frequency \( f \) is given by \( f = \frac{\omega}{2\pi} \). With \( \omega = 9.9 \), the frequency is \( f = \frac{9.9}{2\pi} \approx 1.576 \text{ Hz} \).

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle that describes how springs behave when forces are applied to them. It states that the force, \( F \), exerted by a spring is directly proportional to the amount it is stretched or compressed, which is described by the equation: \[ F = kx \] Here,

• \( F \) is the force exerted by the spring (in Newtons).
• \( k \) is the spring constant, which indicates the stiffness of the spring (in N/m).
• \( x \) is the displacement from the equilibrium position (in meters).

In our exercise, we needed to find the spring constant \( k \). Given the mass stretches the spring by 0.1 m under the influence of its weight (0.98 N), we calculated \( k = 9.8 \, \text{N/m} \) using the equation above. This resulted in understanding how much force is needed to stretch the spring by a specific distance.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of spring-mass systems, they help us determine the motion over time. For a spring-mass system, the differential equation is: \[ m\frac{d^2x}{dt^2} + kx = 0 \] where,

• \( m \) is the mass (in kilograms).
• \( \frac{d^2x}{dt^2} \) is the acceleration (second derivative of displacement with respect to time).
• \( k \) is the spring constant.
• \( x \) is the displacement.

In the exercise, substituting \( m = 0.1 \) kg and \( k = 9.8 \) N/m led us to the equation: \[ \frac{d^2x}{dt^2} + 98x = 0 \] This type of equation is known as a second-order linear homogeneous differential equation, a common form encountered in simple harmonic motion problems.

Harmonic Motion

Harmonic motion describes the oscillatory movement seen in systems like spring-mass systems. The solutions to the differential equations governing these systems often form sinusoidal functions indicating periodic motion. The general solution form is:\[ x(t) = A\cos(\omega t) + B\sin(\omega t) \]For our got equation:

• \( x(t) \) represents the position at time \( t \).
• \( A \) and \( B \) are constants determined by initial conditions.
• \( \omega \) is the angular frequency expressed as \( \sqrt{\frac{k}{m}} \).

Using initial conditions from the problem, the solution simplifies to \[ x(t) = -0.2\cos(9.9t) \] indicating the motion starts from 0.2 meters below equilibrium and follows a cosine wave.

Frequency Calculation

The frequency of oscillation is a measure of how many cycles occur in a unit of time, often specified in Hertz (Hz). It is given by the formula:\[ f = \frac{\omega}{2\pi} \]Where,

• \( f \) is the frequency in cycles per second (Hz).
• \( \omega \) is the angular frequency.

From our exercise, we found \( \omega = 9.9 \), so:\[ f = \frac{9.9}{2\pi} \approx 1.576 \, \text{Hz} \]This tells us that the mass on the spring completes approximately 1.576 cycles per second. Frequency is crucial in understanding the dynamic behavior of oscillating systems.