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Chapter 15: Problem 2

A \(0.12 \mathrm{~kg}\) body undergoes simple harmonic motion of amplitude \(8.5 \mathrm{~cm}\) and period \(0.20 \mathrm{~s}\). (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant?

Short Answer

Expert verified
(a) Maximum force is approximately 10.002 N. (b) Spring constant is approximately 118.46 N/m.

Step by step solution

01

Understand the Problem

We are given a body of mass \( m = 0.12 \) kg moving in simple harmonic motion with amplitude \( A = 8.5 \) cm (convert this to meters for calculations: \( A = 0.085 \) m) and period \( T = 0.20 \) s. We need to find two things: (a) the maximum force on the body, and (b) the spring constant \( k \) of the spring causing the motion.
02

Use Maximum Acceleration for Maximum Force

In simple harmonic motion, the maximum force \( F_{max} \) is equivalent to the maximum acceleration \( a_{max} \) times the mass \( m \). Use the formula for maximum acceleration: \( a_{max} = A \omega^2 \), where \( \omega \) is the angular frequency given by \( \omega = \frac{2\pi}{T} \).
03

Calculate Angular Frequency \( \omega \)

Calculate \( \omega = \frac{2\pi}{T} = \frac{2\pi}{0.20} = 10\pi \) rad/s.
04

Calculate Maximum Acceleration \( a_{max} \)

Substitute \( \omega \) and \( A \) into the formula for maximum acceleration: \[ a_{max} = 0.085 \times (10\pi)^2 = 0.085 \times 100\pi^2 \] \( a_{max} \approx 83.35 \text{ m/s}^2 \).
05

Calculate Maximum Force \( F_{max} \)

Use the formula \( F_{max} = m \times a_{max} = 0.12 \times 83.35 = 10.002 \) N. This is the magnitude of maximum force.
06

Calculate Spring Constant \( k \)

For simple harmonic motion, the spring constant \( k \) can be determined using \( k = m\omega^2 \). Thus, \( k = 0.12 \times (10\pi)^2 = 0.12 \times 100\pi^2 \) which yields \( k \approx 118.46 \text{ N/m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant Calculation
To calculate the spring constant in simple harmonic motion, consider the formula used: \[ k = m \omega^2 \] where \( k \) is the spring constant, \( m \) is the mass of the object, and \( \omega \) is the angular frequency. The spring constant essentially measures the stiffness of the spring鈥攈ow much it resists compression or stretching.
  • A higher spring constant indicates a stiffer spring which requires more force to deform a certain amount.
  • In our exercise, using the mass \( m = 0.12 \) kg and angular frequency \( \omega = 10\pi \) rad/s, we find the spring constant: \( k = 0.12 \times (10\pi)^2 \).
  • This calculation gives us \( k \approx 118.46 \) N/m, meaning it requires approximately 118.46 Newtons to change the spring's length by one meter.
Maximum Force in Oscillation
Simple harmonic motion involves forces that reach a maximum magnitude at the peak of the oscillation. The formula to find this maximum force involves the maximum acceleration \( a_{max} \) and the mass \( m \) of the object:\[ F_{max} = m \times a_{max} \]The maximum acceleration is a result of two variables:
  • The maximum displacement from the equilibrium position, or amplitude \( A \).
  • The square of the angular frequency \( \omega \), derived from \( \omega = \frac{2\pi}{T} \).
Substituting these into the equation for maximum acceleration gives us:\[ a_{max} = A \omega^2 \]To determine \( F_{max} \), substitute the values: the mass is \( 0.12 \) kg, \( a_{max} \approx 83.35 \) m/s虏. Hence, \( F_{max} = 0.12 \times 83.35 \approx 10.002 \) N. This force acts when the object is at its extreme positions.
Angular Frequency
Angular frequency \( \omega \) is a critical aspect in the study of oscillations. It's defined by:\[ \omega = \frac{2\pi}{T} \]where \( T \) is the period of the motion, representing the time for one complete cycle. Angular frequency helps us understand how rapidly the body oscillates through its cycle.
  • In the context of simple harmonic motion, a higher angular frequency suggests the object completes more cycles in a given period.
  • For our exercise, with a period \( T = 0.20 \) seconds, the angular frequency becomes \( \omega = \frac{2\pi}{0.20} = 10\pi \) rad/s.
  • This tells us how fast the oscillations are occurring in terms of radians per second.
Understanding \( \omega \) is essential for calculations involving forces and energies within systems undergoing simple harmonic motion.

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Most popular questions from this chapter

A block weighing \(10.0 \mathrm{~N}\) is attached to the lower end of a vertical spring \((k=200.0 \mathrm{~N} / \mathrm{m})\), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of \(2.00 \mathrm{~J}\) as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

A vertical spring stretches \(9.6 \mathrm{~cm}\) when a \(1.3 \mathrm{~kg}\) block is hung from its end. (a) Calculate the spring constant. This block is then displaced an additional \(5.0 \mathrm{~cm}\) downward and released from rest. Find the (b) period, (c) frequency, (d) amplitude, and (e) maximum speed of the resulting SHM.

A \(55.0 \mathrm{~g}\) block oscillates in SHM on the end of a spring with \(k=1500 \mathrm{~N} / \mathrm{m}\) according to \(x=x_{m} \cos (\omega t+\phi) .\) How long does the block take to move from position \(+0.800 x_{m}\) to (a) position \(+0.600 x_{m}\) and (b) position \(-0.800 x_{m} ?\)

A block rides on a piston (a squat cylindrical piece) that is moving vertically with simple harmonic motion. (a) If the SHM has period \(1.0 \mathrm{~s}\), at what amplitude of motion will the block and piston separate? (b) If the piston has an amplitude of \(5.0 \mathrm{~cm}\), what is the maximum frequency for which the block and piston will be in contact continuously?

The suspension system of a \(2000 \mathrm{~kg}\) automobile "sags" \(10 \mathrm{~cm}\) when the chassis is placed on it. Also, the oscillation amplitude decreases by \(50 \%\) each cycle. Estimate the values of (a) the spring constant \(k\) and (b) the damping constant \(b\) for the spring and shock absorber system of one wheel, assuming each wheel supports \(500 \mathrm{~kg}\).
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