/*! 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} Q3Q The acceleration of a (t) partic... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The acceleration of a (t) particle undergoing SHM is graphed in Fig. 15-21. (a) Which of the labeled points corresponds to the particle at-xm? (b) At point 4, is the velocity of the particle positive, negative, or zero? (c) At point5, is the particle at -xm, or at +xm, at 0, between and, or between 0 and +xm?

Short Answer

Expert verified
  1. Point 2 corresponds to the particle at -xm.
  2. The velocity of particle is positive at point 4
  3. The particle is between 0 at xmat point 5.

Step by step solution

01

The given data 

The graph of acceleration versus time for a particle undergoing SHM is given.

02

Understanding the concept of SHM of a particle

From the given graph, we can determine the position of the particle. We use the direction of acceleration according to the position of the particle. The velocity of the particle is positive, negative or zero can be determined from it.

Formula:

The acceleration of the body in SHM, a=-Ó¬2x (i)

03

Step 3: Calculation of the point that corresponds to the particle at -xm

a)

We know that in acceleration, SHM is always negative corresponding equation (i). When the particle moves away from mean position and acceleration is positive, the particle moves towards mean position.

Here particle is at point 2; acceleration is positive that is the particle is moving toward mean position. Also, the acceleration has maximum value at point 2.

Hence, point 2 corresponds to the particle at -xm.

04

Calculation for the velocity at point 4

b)

At point 4, the particle is at mean position because acceleration is zero. It came from -xmand is moving towards +xmso the velocity is positive.

05

Calculation of the position of particle at point 5

c)

At point 5, the particle is moving from mean position towards+xmand having negative acceleration.

Hence the particle is between 0 and + xm.

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!

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

Question: A performer seated on a trapeze is swinging back and forth with a period of 8.85s . If she stands up, thus raising the center of mass of the trapeze+ performer system by 35.0 cm , what will be the new period of the system? Treat trapeze+ performer as a simple pendulum.

A torsion pendulum consists of a metal disk with a wire running through its center and soldered in place. The wire is mounted vertically on clamps and pulled taut. 15-58a Figuregives the magnitude τof the torque needed to rotate the disk about its center (and thus twist the wire) versus the rotation angle θ. The vertical axis scale is set by τs=4.0×10-3N.m.=.The disk is rotated to θ=0.200rad and then released. Figure 15-58bshows the resulting oscillation in terms of angular position θversus time t. The horizontal axis scale is set by ts=0.40s. (a) What is the rotational inertia of the disk about its center? (b) What is the maximum angular speedof dθ/dtthe disk? (Caution: Do not confuse the (constant) angular frequency of the SHM with the (varying) angular speed of the rotating disk, even though they usually have the same symbol. Hint: The potential energy U of a torsion pendulum is equal to 12kθ2, analogous to U=12kx2for a spring.)

Question: In Figure, the block has a mass of 1.50kgand the spring constant is800 N/m. The damping force is given by -b(dx/dt), where b = 230 g/s. The block is pulled down 12.0 cmand released.

  1. Calculate the time required for the amplitude of the resulting oscillations to fall to one-third of its initial value.
  2. How many oscillations are made by the block in this time?

Question: A rectangular block, with face lengths a = 35 cmand b = 45 cm, is to be suspended on a thin horizontal rod running through a narrow hole in the block. The block is then to be set swinging about the rod like a pendulum, through small angles so that it is in SHM. Figure shows one possible position of the hole, at distance rfrom the block’s center, along a line connecting the center with a corner.

  1. Plot the period of the pendulum versus distance ralong that line such that the minimum in the curve is apparent.
  2. For what value of rdoes that minimum occur? There is actually a line of points around the block’s center for which the period of swinging has the same minimum value.
  3. What shape does that line make?

What is the maximum acceleration of a platform that oscillates at amplitude 2.20 cmand frequency 6.60 Hz?

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.