/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A \(1.50-\mathrm{kg}\) mass on a... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(1.50-\mathrm{kg}\) mass on a spring has displacement as a function of time given by the equation $$x(t)=(7.40 \mathrm{cm}) \cos \left[\left(4.16 \mathrm{s}^{-1}\right) t-2.42\right]$$ Find (a) the time for one complete vibration; (b) the force constant of the spring; (c) the maximum speed of the mass; (d) the maximum force on the mass; (e) the position, speed, and acceleration of the mass at \(t=1.00 \mathrm{s} ;(\mathrm{f})\) the force on the mass at that time.

Short Answer

Expert verified
(a) 1.51 s, (b) 26.0 N/m, (c) 0.308 m/s, (d) 1.92 N, (e) 0.065 m, 0.109 m/s, -0.374 m/s², (f) -0.561 N.

Step by step solution

01

Identify the Angular Frequency

The displacement equation for the mass is given as \( x(t) = (7.40 \text{ cm}) \cos[(4.16 \text{ s}^{-1})t - 2.42] \). The angular frequency \( \omega \) is the coefficient of \( t \) inside the cosine function, which is \( 4.16 \text{ s}^{-1} \).
02

Compute the Time for One Complete Vibration

The time for one complete vibration, or period \( T \), is given by the formula \( T = \frac{2\pi}{\omega} \). Plugging in \( \omega = 4.16 \text{ s}^{-1} \), we find \( T = \frac{2\pi}{4.16} \approx 1.51 \text{ s} \).
03

Calculate the Force Constant of the Spring

The force constant \( k \) is related to the angular frequency by the equation \( \omega = \sqrt{\frac{k}{m}} \). Rearranging gives \( k = m\omega^2 \). With \( m = 1.50 \text{ kg} \) and \( \omega = 4.16 \text{ s}^{-1} \), we find \( k = 1.50 \times (4.16)^2 \approx 26.0 \text{ N/m} \).
04

Determine the Maximum Speed of the Mass

The maximum speed \( v_{max} \) occurs when the displacement is zero and is given by \( v_{max} = A \omega \), where \( A = 7.40 \text{ cm} = 0.074 \text{ m} \). So, \( v_{max} = 0.074 \times 4.16 \approx 0.308 \text{ m/s} \).
05

Find the Maximum Force on the Mass

The maximum force \( F_{max} \) is given by \( F_{max} = kA \). Using \( k = 26.0 \text{ N/m} \) and \( A = 0.074 \text{ m} \), we find \( F_{max} = 26.0 \times 0.074 \approx 1.92 \text{ N} \).
06

Compute Position, Speed, and Acceleration at \( t = 1.00 \text{ s} \)

Plug \( t = 1.00 \) into the equations:- Position: \( x(1.00) = 0.074 \cos((4.16 \times 1.00) - 2.42) \approx 0.065 \text{ m} \).- Speed: \( v(t) = -\omega A \sin((\omega t) - 2.42) \). Thus, \( v(1.00) = -0.308 \sin((4.16 \times 1.00) - 2.42) \approx 0.109 \text{ m/s} \).- Acceleration: \( a(t) = -\omega^2 A \cos((\omega t) - 2.42) \). So, \( a(1.00) = -(4.16)^2 \times 0.074 \cos((4.16 \times 1.00) - 2.42) \approx -0.374 \text{ m/s}^2 \).
07

Calculate the Force on the Mass at \( t = 1.00 \text{ s} \)

The force \( F(t) \) on the mass at any time is \( m \cdot a(t) \). With \( a(1.00) = -0.374 \text{ m/s}^2 \), the force is \( F(1.00) = 1.50 \times -0.374 \approx -0.561 \text{ N} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Frequency
Angular frequency, often symbolized as \( \omega \), is a key element in simple harmonic motion (SHM) and it tells us how quickly an object oscillates. In the context of the given exercise, the angular frequency is positioned as the multiplier of time \( t \) in the cosine function of the displacement equation. Specifically, it is \( 4.16 \, \text{s}^{-1} \).

This constant provides insight into how many radians the oscillation completes per second. To find the time it takes to complete one full cycle, we use the period \( T \) formula: \[T = \frac{2\pi}{\omega}\] Plugging in \( \omega = 4.16 \, \text{s}^{-1} \), the period is computed as approximately 1.51 seconds. This means that in this SHM scenario, the mass completes one full oscillation in about 1.51 seconds.
Spring Constant
The spring constant, represented as \( k \), signifies the stiffness of a spring in simple harmonic motion. It is determined from the relationship between angular frequency and mass of the oscillating object. The equation linking these parameters is:
  • \( \omega = \sqrt{\frac{k}{m}} \)
Rearranging this for \( k \) gives us:\[k = m\omega^2\]Given \( m = 1.50 \, \text{kg} \) and \( \omega = 4.16 \, \text{s}^{-1} \), the spring constant is calculated as \( k = 1.50 \times (4.16)^2 \approx 26.0 \, \text{N/m} \).

This reveals that the spring requires 26.0 Newtons of force to stretch it by one meter, indicating how rigid or flexible the spring is during oscillations.
Maximum Speed
In simple harmonic motion, the maximum speed \( v_{\text{max}} \) of the oscillating object is a crucial aspect to grasp, as it relates to how quickly the object moves. This speed is achieved when the object's displacement is zero. The formula for maximum speed is:
  • \( v_{\text{max}} = A \omega \)
where \( A \) is the amplitude of oscillation and \( \omega \) is the angular frequency.

For the exercise at hand, with \( A = 0.074 \, \text{m} \) (converted from 7.40 cm) and \( \omega = 4.16 \, \text{s}^{-1} \), the maximum speed is \( v_{\text{max}} = 0.074 \times 4.16 \approx 0.308 \, \text{m/s} \).

This means that the mass achieves its quickest motion speed of 0.308 meters per second as it passes through the equilibrium position in its oscillatory path.
Displacement Function
The displacement function describes how the position of an object in simple harmonic motion varies over time. In our specific equation, it is:\[x(t) = (7.40 \, \text{cm}) \cos[(4.16 \, \text{s}^{-1})t - 2.42]\]This function uses cosine to show the oscillation profile, where:
  • The amplitude \( A = 7.40 \, \text{cm} \) indicates the maximum distance from the equilibrium.
  • The term \( 4.16 \, \text{s}^{-1} \) corresponds to the angular frequency.
  • \(-2.42\) is the phase shift altering the starting point of the oscillation.
Plugging in a particular time, like \( t=1.00 \, \text{s} \), gives us the precise position at that moment, \( x(1.00) \approx 0.065 \, \text{m}\).

This demonstrates how the displacement at any point in time can be determined, revealing the mass's precise location as it swings back and forth.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(1.50-\mathrm{kg}\) ball and a \(2.00-\mathrm{kg}\) ball are glued together with the lighter one below the heavier one. The upper ball is attached to a vertical ideal spring of force constant \(165 \mathrm{N} / \mathrm{m},\) and the system is vibrating vertically with amplitude 15.0 \(\mathrm{cm} .\) The gine connecting the balls is old and weak, and it suddenly comes loose when the balls are at the lowest position in their motion. (a) Why is the glue more likely to fail at the lowest point than at any other point in the motion? (b) Find the amplitude and frequency of the vibrations after the lower ball has come loose.

A harmonic oscillator consists of a \(0.500-\mathrm{kg}\) mass attached to an ideal spring with force constant 140 \(\mathrm{N} / \mathrm{m}\) . Find (a) the period; (b) the frequency; (c) the angular frequency of the oscillations.

Show that the expression for the period of a physical penduhm reduces to that of a simple pendulum if the physical pendulum consists of a particle with mass \(m\) on the end of a massless string of length \(L .\)

A uniform rod of length \(L\) oscillates through small angles about a point a distance \(x\) from its center. (a) Prove that its angular frequency is \(\sqrt{g x /\left[\left(L^{2} / 12\right)+x^{2}\right] .(b)}\) Show that its maximum angular frequency occurs when \(x=L / \sqrt{12}\) . (c) What is the length of the rod if the maximum angular frequency is 2\(\pi \mathrm{rad} / \mathrm{s} ?\)

A proud deep-sea fisherman hangs a \(65.0-\mathrm{kg}\) fish from an ideal spring having negligible mass. The fish stretches the spring 0.120 \(\mathrm{m}\) m. Find the force constant of the spring. The fish is now pulled down 5.00 \(\mathrm{cm}\) and released. (b) What is the period of oscillation of the fish? (c) What is the maximum speed it will reach?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.