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(II) A balsa wood block of mass 55 $\mathrm{g}$ floats on a lake, bobbing up and down at a frequency of 3.0 $\mathrm{Hz}$ (a) What is the value of the effective spring constant of the water? (b) A partially filled water bottle of mass 0.25 $\mathrm{kg}$ and almost the same size and shape of the balsa block is tossed into the water. At what frequency would you expect the bottle to bob up and down? Assume SHM.

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a. $$19.54 \mathrm{N} / \mathrm{m}$$b. $$1.4 \mathrm{Hz}$$

Physics 101 Mechanics

Chapter 14

Oscillators

Motion Along a Straight Line

Motion in 2d or 3d

Periodic Motion

Cornell University

Hope College

University of Winnipeg

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.

02:53

A balsa wood block of mass…

02:05

02:14

01:18

A block of wood of mass 55…

03:37

A block ofwood of mass $4 …

02:19

A 320 -kg wooden raft floa…

03:13

A 320-kg wooden raft float…

06:22

04:58

A 350-kg wooden raft float…

01:11

A wooden block of mass $0-…

01:55

(II) A fisherman's sc…

02:23

A fisherman's scale s…

01:25

A large rectangular barge …

01:35

02:58

A 50-cm-long spring is sus…

02:04

III A 50-cm-long spring is…

02:01

A spring is hung from the …

01:50

Figure $P 14.59$ shows two…

so for party were first trying to find thespian constant. We know the frequency is equal to one over to pie times the square root of the spring constant divided by the mass. Therefore, the spring constant is equal to four pi squared, uh, half square, the frequency squared times the mass. And at this point, we can solve, So K is gonna be equal to four pi squared, Um, times three hurts, quantity squared times the mass of 0.55 kilograms. And this is giving us 19.54 a newtons per meter. So for party, this would be our final answer. And then for part B, we want to find the frequency such that the masses increased from 0.55 kilograms, 2.66 kilograms. So the frequency would again be equal to one over to pie times, the square root of the spring constant, which we found to be 19.4 newtons per meter divided by the mass, the new mass, which would be 0.66 kilograms. And we find that the frequency is going to be equal to 1.4 hertz approximately so this would be our final answer for part B. That is the end of the solution. Thank you for watching

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