/*! 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 76 A \(0.0200-\mathrm{kg}\) bolt mo... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(0.0200-\mathrm{kg}\) bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{c})\) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;(\mathrm{d})\) the speed of the bolt when \(x=-0.180 \mathrm{m} .\)

Short Answer

Expert verified
(a) -0.120 m; (b) 0.0421 N, positive direction; (c) 0.577 s; (d) 0.642 m/s.

Step by step solution

01

Calculate the Angular Frequency

The angular frequency \( \omega \) is given by the formula \( \omega = \frac{2\pi}{T} \) where \( T \) is the period. Here, \( T = 1.500 \, \mathrm{s} \). So,\[ \omega = \frac{2\pi}{1.500} \approx 4.1888 \, \mathrm{rad/s} \]
02

Determine the Displacement at t=0.500 s

The displacement of the bolt is described by the equation \( x(t) = A \cos(\omega t) \). Here, \( A = 0.240 \, \mathrm{m} \) and the initial displacement is \(+0.240 \, \mathrm{m} \). Thus, at \( t = 0.500 \, \mathrm{s} \),\[ x(0.500) = 0.240 \cos(4.1888 \times 0.500) \approx 0.240 \cos(2.0944) \approx -0.120 \, \mathrm{m} \]
03

Calculate the Force Acting on the Bolt at t=0.500 s

The force \( F \) in SHM is given by \( F = -m\omega^2 x \). We have \( m = 0.0200 \, \mathrm{kg} \), \( \omega = 4.1888 \, \mathrm{rad/s} \), and \( x = -0.120 \, \mathrm{m} \). Therefore,\[ F = -(0.0200)(4.1888)^2(-0.120) \approx 0.0421 \, \mathrm{N} \] in the positive direction.
04

Find the Minimum Time to Move to x=-0.180 m

To find this time, we solve the equation \( x(t) = A \cos(\omega t) = -0.180 \, \mathrm{m} \).\[ -0.180 = 0.240 \cos(4.1888 t) \]So, \[ \cos(4.1888 t) = -\frac{0.180}{0.240} = -0.75 \]\[ 4.1888 t = \cos^{-1}(-0.75) \approx 2.418 \]\[ t = \frac{2.418}{4.1888} \approx 0.577 \, \mathrm{s} \]
05

Calculate the Bolt's Speed When x=-0.180 m

The speed in SHM is given by \( v = \omega \sqrt{A^2 - x^2} \). We have \( A = 0.240 \, \mathrm{m} \), \( x = -0.180 \, \mathrm{m} \), and \( \omega = 4.1888 \, \mathrm{rad/s} \). Thus,\[ v = 4.1888 \sqrt{(0.240)^2 - (-0.180)^2} \approx 4.1888 \sqrt{0.0576 - 0.0324} \approx 4.1888 \times 0.153 \approx 0.642 \, \mathrm{m/s} \]

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, represented by \( \omega \), is a fundamental characteristic in Simple Harmonic Motion (SHM). It describes how quickly the object oscillates back and forth.
The formula to calculate angular frequency is given by \( \omega = \frac{2\pi}{T} \), where \( T \) is the period, the time it takes to complete one full cycle of the motion. For our bolt, the period \( T \) is 1.500 seconds. Plugging this into our formula, we calculate the angular frequency as \( \omega \approx 4.1888 \, \text{rad/s} \).
This value of \( \omega \) signifies that the bolt completes the equivalent of \( 4.1888 \) radians per second during its oscillation. Angular frequency is essential because it ties together time, displacement, and force in SHM, providing a complete picture of the motion's characteristics.
Displacement Calculation
In SHM, the displacement of an object from its mean position at any time is given by the equation \( x(t) = A \cos(\omega t) \), where \( A \) is the amplitude, or the maximum displacement from the mean. Here, \( A = 0.240 \, \text{m} \).
To find the displacement of the bolt at a specific time, say \( t = 0.500 \, \text{s} \), we substitute \( t \) and the angular frequency \( \omega \) into our displacement equation. This gives \( x(0.500) = 0.240 \cos(4.1888 \times 0.500) \approx -0.120 \, \text{m} \).
The negative displacement indicates the bolt is on the opposite side of its starting position, moving towards the equilibrium point. Understanding displacement helps visualize how far and in which direction the object has moved at any given time during its oscillation.
Force in SHM
In SHM, the force acting on an object is always directed towards the equilibrium position and is proportional to the displacement from it. This force can be found using Hooke's Law for SHM, expressed mathematically as \( F = -m\omega^2 x \).
For our bolt, with mass \( m = 0.0200 \, \text{kg} \), angular frequency \( \omega = 4.1888 \, \text{rad/s} \), and displacement \( x = -0.120 \, \text{m} \), we calculate the force as \( F \approx 0.0421 \, \text{N} \).
The positive sign in the result indicates that the force is in the positive direction, meaning it is acting towards the original position of equilibrium. Grasping how force is calculated in SHM aids in understanding the restoring nature of SHM, where the force aims to bring the object back to the center of motion.
Velocity in SHM
The velocity of an object in SHM can tell us a lot about its energy state and speed at different positions during the cycle. It is given by the equation \( v = \omega \sqrt{A^2 - x^2} \).
For the bolt, with amplitude \( A = 0.240 \, \text{m} \) and displacement \( x = -0.180 \, \text{m} \), the velocity at this position can be calculated. Using the already found angular frequency \( \omega = 4.1888 \, \text{rad/s} \), the velocity becomes \( v \approx 0.642 \, \text{m/s} \).
This indicates how fast the bolt is moving at that point and helps us understand the dynamic motion at various stages of the SHM process. By determining the velocity, one can analyze not just the speed, but also gain insights into the kinetic energy changes as the object oscillates.

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

An \(85.0-\mathrm{kg}\) mountain climber plans to swing down, starting from rest, from a ledge using a light rope 6.50 \(\mathrm{m}\) long. He holds one end of the rope, and the other end is tied higher up on a rock face. Since the ledge is not very far from the rock face, the rope makes a small angle with the vertical. At the lowest point of his swing, he plans to let go and drop a short distance to the ground. (a) How long after he begins his swing will the climber first reach his lowest point? (b) If he missed the first chance to drop off, how long after first beginning his swing will the climber reach his lowest point for the second time?

The two pendulums shown in Fig 13.34 each consist of a uniform solid ball of mass \(M\) supported by a massless string, but the ball for pendulum \(A\) is very tiny while the ball for pendulum \(B\) is much larger. Find the period of each pendulum for small displacements. Which ball takes longer to complete a swing?

You want to construct a pendulum with a period of 4.00 \(\mathrm{s}\) at a location where \(g=\) 9.80 \(\mathrm{m} / \mathrm{s}^{2} .\) (a) What is the length of a simple pendulum having this period? (b) Suppose the pendulum must be mounted in a case that is not more than 0.50 \(\mathrm{m}\) high. Can you devise a pendulum having a period of 4.00 s that will satisfy this requirement?

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 thrill-seeking cat with mass 4.00 \(\mathrm{kg}\) is attached by a harness to an ideal spring of negligible mass and oscillates verticallyin SHM. The amplitude is \(0.050 \mathrm{m},\) and at the highest point of the motion the spring has its natural unstretched length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum of these three energies when the cat is (a) at its highest point; ( \((b)\) at its lowest point; (c) at its equilibrium position.

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.