Solving for Velocity and Describing Motion of a Spring-Mass System

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Homework Help Overview

The discussion revolves around solving for the velocity and describing the motion of a spring-mass system, specifically focusing on the dynamics of a physical pendulum and the application of the formula for the period of oscillation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants share their attempts to calculate the length of a pendulum using the period formula, but express confusion over incorrect results. There are mentions of the moment of inertia and the nature of the system as a physical pendulum. One participant raises a new problem regarding the velocity of an object attached to a spring, indicating uncertainty about the starting point and relevant equations.

Discussion Status

The discussion includes various attempts to solve the initial problem, with some participants providing clues and hints without reaching a consensus. There is an ongoing exploration of different aspects of the spring-mass system, including the transition to a new problem about velocity, which remains unresolved.

Contextual Notes

Participants are working under constraints typical of homework assignments, such as specific formulas and conditions related to the physical properties of the systems being analyzed. There is also a note about not bumping threads until a certain time has passed.

riseofphoenix
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#7.
Number7.png


This is what i tried to do... but I got it wrong :( Please help

T ≈ 2π√(L/g)

1.31 = 2π√(L/9.81)
1.31/2π = √(L/9.81)
2.057 = √(L/9.81)
2.057² = L/9.81
4.2343 = L/9.81
4.2343(9.81) = L
41.54/1000 = L
0.0415 = L

"INCORRECT: Your response differs from the correct answer by more than 100%. "
 
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Wow where is everybody?
 
Hello?
 
riseofphoenix said:
#7.
Number7.png


This is what i tried to do... but I got it wrong :( Please help

T ≈ 2π√(L/g)

1.31 = 2π√(L/9.81)
1.31/2π = √(L/9.81)
2.057 = √(L/9.81)
2.057² = L/9.81
4.2343 = L/9.81
4.2343(9.81) = L
41.54/1000 = L
0.0415 = L

"INCORRECT: Your response differs from the correct answer by more than 100%. "
One clue is the mention of the moment of inertia.

Another is that it's a physical pendulum.

Also,
Don't bump your thread until it's been here for 24 hours.​
 
SammyS said:
One clue is the mention of the moment of inertia.

Another is that it's a physical pendulum.

Also,
Don't bump your thread until it's been here for 24 hours.​

I =T2MgD/4π2
I = [(1.312)(0.246)(9.81)(18)]/39.478
I = 74.545/39.478
I = 1.8882

"INCORRECT"

-___-
AKFJDJFDAKLNDJK
 
riseofphoenix said:
I =T2MgD/4π2
I = [(1.312)(0.246)(9.81)(18)]/39.478
I = 74.545/39.478
I = 1.8882

"INCORRECT"

-___-
AKFJDJFDAKLNDJK
Convert centimeters to meters.
 
SammyS said:
Convert centimeters to meters.

Oh...

0.0188

"CORRECT"

-.- The day I graduate from Webassign...

Thank you...

What about this last one? I have no idea where to start/what equation to use.

#3. A 40.0-g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 20.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
m/s
 
riseofphoenix said:
Oh...

0.0188

"CORRECT"

-.- The day I graduate from Webassign...

Thank you...

What about this last one? I have no idea where to start/what equation to use.

#3. A 40.0-g object is attached to a horizontal spring with a force constant of 10.0 N/m and released from rest with an amplitude of 20.0 cm. What is the velocity of the object when it is halfway to the equilibrium position if the surface is frictionless?
m/s
Describe the motion of the object after being released.
 

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