Constant of Integration when calculating displacement

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Homework Help Overview

The discussion revolves around the integration of a velocity function for a pendulum bob in simple harmonic motion, specifically addressing the constant of integration when calculating displacement. The velocity is given as v = 1.26sin(2πt), and the participants explore the implications of the constant of integration in the context of initial conditions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the integration of the velocity function and the resulting displacement equation, questioning the significance of the constant of integration. They explore whether it can be ignored and how initial conditions affect the calculation.

Discussion Status

There is an active exploration of the role of initial conditions in determining the constant of integration. Some participants provide insights into the behavior of the pendulum at t=0, while others express confusion regarding the relationship between velocity and displacement in simple harmonic motion.

Contextual Notes

Participants are navigating assumptions about the initial position and velocity of the pendulum, with some suggesting that the common understanding of simple harmonic motion may not align with the original poster's assumptions.

RubiksMelia
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Homework Statement


The speed of a pendulum bob moving in simple harmonic motion is given by v = 1.26sin(2πt) where v is in m/s and t is time in seconds.

Homework Equations


s = ∫ v dt

The Attempt at a Solution



v = 1.26sin(2πt)

Integrating v yields

s = -0.2cos(2πt) + c

and solving for c where s, t = 0 yields c = 0.2 (as cos 2πt = 1 when t = 0), meaning that s = 0.2 - 0.2cos(2πt)

The answer in the back of the textbook seems to ignore this fact, simply stating that the displacement is equal to -0.2cos(2πt).

My question is this: does the constant of integration always matter when integrating like this to find displacements and velocities? Can it be "ignored"? Is the back of my textbook lying to me about the answer? :wink:
 
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RubiksMelia said:
and solving for c where s, t = 0 yields c = 0.2 (as cos 2πt = 1 when t = 0), meaning that s = 0.2 - 0.2cos(2πt)
You can see at t=0, the pendulum has zero velocity, meaning it is at one of the extremities. So, at t=0, the displacement should be maximum.
 
Last edited:
cnh1995 said:
You can see at t=0, the pendulum has zero velocity, meaning it is at one of the extremities. So, at t=0, the displacement should be maximum.

But, at t = 0, shouldn't the pendulum be at rest in the center (rather than the left or right extremity)?
 
The constant of integration matters and it is calculated by the initial conditions.
You seem to take as initial condition s(0)=0, but this is not correct and the reason is as cnh1995's post explains.
 
RubiksMelia said:
But, at t = 0, shouldn't the pendulum be at rest in the center (rather than the left or right extremity)?
In SHM, the pendulum is at the center when its velocity is maximum. Its velocity goes on decreasing till the extremity, where it becomes zero. In SHM, t=0 does not mean the pendulum is at the center.
 
RubiksMelia said:
But, at t = 0, shouldn't the pendulum be at rest in the center (rather than the left or right extremity)?
. This means that s(0)=0, v(0)=0 but this means that the pendulum will stay at center and do nothing (the force is zero at the center).
 
cnh1995 said:
In SHM, the pendulum is at the center when its velocity is maximum. Its velocity goes on decreasing till the extremity, where it becomes zero. In SHM, t=0 does not mean the pendulum is at the center.
Thanks! I should have had a look at simple harmonic motion. Silly me for tackling a question before knowing what everything meant. Thanks to all for the help.
 

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