Problem 36 - College Physics

Question

A mass of \(0.20 \mathrm{~kg}\) on the end of a spring oscillates with a period of \(0.45 \mathrm{~s}\) and an amplitude of \(0.15 \mathrm{~m} .\) Find (a) the velocity when it passes the equilibrium point, (b) the total energy of the system, (c) the spring constant, and (d) the maximum acceleration of the mass.

Step-by-Step Solution

Answer

(a) 2.09 m/s, (b) 0.44 J, (c) 39.0 N/m, (d) 29.2 m/s².

Step 1: Calculate Angular Frequency

The angular frequency \( \omega \) of the spring-mass system is given by the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of oscillation. Substituting the given period, \( T = 0.45 \) s, we have: \[ \omega = \frac{2\pi}{0.45} \approx 13.96 \, \text{rad/s}. \]

Step 2: Velocity at Equilibrium Point

At the equilibrium point, the velocity of a mass in simple harmonic motion is maximum, calculated by the formula \( v_{max} = \omega A \), where \( A \) is the amplitude. Substituting the given amplitude and angular frequency: \[ v_{max} = 13.96 \times 0.15 \approx 2.09 \, \text{m/s}. \]

Step 3: Calculate Total Energy

The total mechanical energy \( E \) in a spring-mass system is given by \( E = \frac{1}{2} m \omega^2 A^2 \). Substituting the known values (\( m = 0.20 \, \text{kg} \), \( \omega \approx 13.96 \, \text{rad/s} \), and \( A = 0.15 \, \text{m} \)): \[ E = \frac{1}{2} \times 0.20 \times (13.96)^2 \times (0.15)^2 \approx 0.44 \, \text{J}. \]

Step 4: Calculate Spring Constant

The spring constant \( k \) is related to the angular frequency and mass by \( \omega = \sqrt{\frac{k}{m}} \). Rearranging this gives \( k = m \omega^2 \). Substituting for \( m \) and \( \omega \), we find: \[ k = 0.20 \times (13.96)^2 \approx 39.0 \, \text{N/m}. \]

Step 5: Maximum Acceleration

The maximum acceleration \( a_{max} \) occurs at maximum displacement and is given by \( a_{max} = \omega^2 A \). Using the established angular frequency and amplitude, we find: \[ a_{max} = (13.96)^2 \times 0.15 \approx 29.2 \, \text{m/s}^2. \]

Key Concepts

Understanding Angular Frequency

In simple harmonic motion, angular frequency, denoted by \( \omega \, \), plays a crucial role. It describes how quickly an object oscillates back and forth. The formula to calculate angular frequency is \[ \omega = \frac{2\pi}{T} \]where

• \( \omega \): Angular frequency (in radians per second)
• \( T \): Period of oscillation (time taken for one complete cycle)

In our example, the period \( T \) is given as 0.45 seconds. Calculating, we find that \( \omega \approx 13.96 \, \) rad/s. This tells us that every second, the mass completes about 14 radians of its path. Angular frequency is crucial for determining maximum velocity and acceleration in oscillatory systems. It's what connects the motion of the mass to the characteristics of the spring.

Spring Constant: The Measure of Stiffness

The spring constant, denoted by \( k \, \), is a measure of the stiffness of a spring. It tells us how much force is needed to stretch or compress the spring by a unit length. It's a fundamental property in Hooke's Law. To find the spring constant, we use the relationship: \[ k = m \omega^2 \]where:

• \( k \): Spring constant (N/m)
• \( m \): Mass attached to the spring (kg)
• \( \omega \): Angular frequency (rad/s)

Substituting the values \( m = 0.20 \, \) kg and \( \omega = 13.96 \, \) rad/s, we get \( k \approx 39.0 \, \) N/m. This indicates a relatively stiff spring, as a fairly substantial force is needed to change its length.

Total Mechanical Energy in Oscillatory Motion

Mechanical energy in a system is always conserved, assuming no external forces like friction act on it. For a simple harmonic oscillator, the mechanical energy consists of potential energy (when the spring is stretched or compressed) and kinetic energy (when the mass is moving). The formula for total mechanical energy \( E \, \) is: \[ E = \frac{1}{2} m \omega^2 A^2 \]where:

• \( E \): Total energy (Joules)
• \( A \): Amplitude of motion (m)

In our calculation, using \( m = 0.20 \, \) kg, \( \omega = 13.96 \, \) rad/s, and \( A = 0.15 \, \) m, we found \( E \approx 0.44 \, \) J. This is the energy the mass exchanges between potential and kinetic forms as it oscillates, ping ponging back and forth due to the restoring force of the spring.

Maximum Acceleration in Simple Harmonic Motion

Maximum acceleration in simple harmonic motion occurs at the points of maximum displacement. It's directly tied to the angular frequency and amplitude of the oscillation. The formula to calculate maximum acceleration \( a_{max} \, \) is: \[ a_{max} = \omega^2 A \]where:

• \( a_{max} \): Maximum acceleration (m/s²)
• \( \omega \): Angular frequency (rad/s)
• \( A \): Amplitude of oscillation (m)

Substituting in \( \omega = 13.96 \, \) rad/s and \( A = 0.15 \, \) m, the maximum acceleration comes out to \( a_{max} \approx 29.2 \, \) m/s². This peak acceleration tells us how strong the forces acting on the mass are at the extreme points of its path, giving an idea of the dynamic nature of the motion.