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(II) $\mathrm{A} .3 .35$ -kg mass at the end of a spring oscillates 2.5 times per second with an amplitude of 0.15 $\mathrm{m} .$ Determine $(a)$ the velocity when it passes the equilibrium point, (b) the velocity when it is 0.10 $\mathrm{m}$ from equilibrium, (c) the total energy of the system, and $(d)$ the equation describing the motion of the mass, assuming that at $t=0, x$ was a maximum.

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a. the linear velocity of the object is 2.4 $\mathrm{m} / \mathrm{s}$b. the linear velocity of the object when it is 0.10 $\mathrm{m}$ for equilibrium is 1.73 $\mathrm{m} / \mathrm{s}$c. the energy of the system is 0.97 $\mathrm{J}$d. the equation of the motion of the wave is 0.15 $\cos (5 \pi t)$

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

University of Washington

Simon Fraser University

Hope College

Lectures

04:01

2D kinematics is the study of the movement of an object in two dimensions, usually in a Cartesian coordinate system. The study of the movement of an object in only one dimension is called 1D kinematics. The study of the movement of an object in three dimensions is called 3D kinematics.

02:18

In physics, an oscillation is the repetitive variation, typically in time, of some measure about a central value or between two or more different states. The oscillation may be periodic or aperiodic.

0:00

A 0.25-kg mass at the end …

04:35

07:25

d) A 0.35 kg mass is attac…

03:50

$\bullet$ A mass of 0.20 $…

04:49

(II) A $0.60-\mathrm{kg}$ …

02:48

A mass of $0.20 \mathrm{~k…

03:49

(II) An object with mass 3…

04:23

A mass-spring system oscil…

02:09

An object-spring system os…

02:42

(II) An object with mass 2…

04:22

(II) A vertical spring wit…

03:21

An object with mass 2.7 kg…

04:24

A mass $m=4.5 \mathrm{kg}$…

01:43

A 0.85-kg mass attached to…

02:38

III The position of a 50 g…

04:03

A 2.0-kg object is attache…

04:19

$\bullet\bullet$ A mass-sp…

So for a party, we want to find the maximum velocity. This will be equal to the angular velocity times the amplitude and this would be equal to two pi times the frequency times, the amplitude, Uh, now we can solve. So this would be to pie times the frequency of 2.5 hertz and then times the amplitude of 0.15 meters, and we find that the maximum velocity equals 2.356 meters per second. Uh, we confined the velocity at equilibrium to then be equal to plus or minus 2.356 meters per second for part B. Ah, velocity. We're trying to find the velocity at a certain displacement. So we consider the velocity is gonna be equal to plus or minus the maximum velocity times the square root of one minus X squared over a square. And this is equaling plus or minus 2.3556 times the square root of one minus 0.10 divided by 0.15 This is squared, and we find that V is gonna be equal to plus or minus 1.76 meters per second. So this would be, uh, the final answer for a part B. My apologies. And then for part C, we want to find the total amount of energy, so this would be the total amount of energy would be equal to 1/2 times the mass times the maximum velocities squared and the sequel in 1/2 times the mass of 0.35 kilograms times the maximum velocity of 2.356 meters per second quantity squared, and we find that the total amount of energy is gonna be equal 2.971 Jules, for a part d we want to find the maximum displacement. Uh, at T equals zero seconds. We know this. Um, if the initial position is going to, um if the initial position is at the maximum displacement, this means that the general equation will be a co sign graph. So this is a co sign function. Ah, and so acts the general formula would be equal to the amplitude of 0.15 meters times co sign of to pi times the frequency of 2.5 hertz times t, and so this is equaling essentially 0.15 meters and then times co signed of five piety, and so this would be our final answer. That is the end of the solution. Thank you for watching.

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