Problem

A 3.2 -kg block is hanging stationary from the en…

03:23

Problem 27 Easy Difficulty

A spring is hung from the ceiling. A $0.450-\mathrm{kg}$ block is then attached to the free end of the spring. When released from rest, the block drops 0.150 $\mathrm{m}$ before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring? (b) Find the
angular frequency of the block's vibrations.

Answer

a) $58.8$ N / m
b) $11.4 $ rad / s

Related Courses

Physics 101 Mechanics

Physics

Chapter 10

Simple Harmonic Motion and Elasticity

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 10

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

Problem 67

Problem 68

Problem 69

Problem 70

Problem 71

Problem 72

Problem 73

Problem 74

Problem 75

Problem 76

Problem 77

Problem 78

Problem 79

Problem 80

Problem 81

Problem 82

Problem 83

Problem 84

Problem 85

Problem 86

Problem 87

Problem 88

Problem 89

Problem 90

Problem 91

Problem 92

Video Transcript

for this question, we're told that a spring is hung from the ceiling. The mass of a block attached to it emits 0.45 kilograms. It was released from rest and the block drop 0.15 meters before momentarily moving back upwards. And it wants to know the spring concert of the spring. So the masses 0.45 But the displacement here is 0.15 um, divided by two. So, uh, this is because the the masses going to also Lee that's the total stretch displacement of the spring. So the masses going toe Ossa late back and forth between half that. Okay, so at this moment, when it comes to rest before it starts going back, back up the force from gravity is equal to the force acting on the spring f sub s. So we just replace these with our conditions. The force of gravity is mass times acceleration of gravity. G. The force of the spring is K times X toasted toe. Find screen constantly. Just simply rearrange this equation and salt for K, we have K is equal to mass times gravity divided by X. Therefore, the spring constant K is equal to 58.8, and the units here would be willing to redraw that eight would be Newtons per meter so he can, Box said. It is our solution to a part. B wants to know the angular frequency of this oscillation. So the angular frequency of the oscillation Omega is by definition equal to the square root of K over in for a simple harmonic oscillating spree. Plugging in those values, we find that this is equal to 11.4 radiance per second tweaking box that is your solution for party.

We have video lessons for 99.27% of the questions in this textbook