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# LO for Physics 101

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## Courtney Cook

on 4 March 2015

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#### Transcript of LO for Physics 101

Unless otherwise indicated some assumptions can always be made
since this is a horizontal mass problem, the force of gravity will not have an effect
this is an ideal system in which friction will have no effect

General formula's
We can then sub in a value of zero for the velocity and determine the time as this would also be the time that the block would be at its maximum acceleration
v(t)=-90sin(3t)=0
t=pi/3
Then we can sub in t=pi/3 into the formula for velocity and determine that the maximum acceleration would be
a(t)=-270cos(3pi/3)
a(t)=270
The time at which the block is at its maximum acceleration can also be determined by setting cos(3t)=1 and solving for t

One must either intuitively understanding that when the block is in the position c it is at a maximum acceleration and simultaneously a velocity of zero or by looking at a graphs of the functions and seeing these values.

Horizontal Mass Spring Question
A 4kg block is attached to a horizontal spring. The spring constant is 36n/m and the block is pulled to its maximum displacement of 30 cm to the right and released at t=0. At what time would the acceleration be at its maximum value and directed towards the right and what will be this value of maximum acceleration?
All of these formulas can be derived by knowing that an oscillating mass spring system is a type of simple harmonic motion and the formula for position of simple harmonic motion is as seen above. Then by realizing that velocity is the derivative of position and acceleration the derivative of velocity, the other formulas can be found.
x(t)=30cos(3t)
v(t)=-90sin(3t)
a(t)=-270cos(3t)
These formulas can be found by understanding that:
the amplitude is the maximum displacement of the spring which in our case is 30 as indicated in the question
the value for phi must be 0 for the displacement formula so that at time t=0 we would be taking the cosine of zero which has a value of one so that the position at t=0 would just be the amplitude which is the case indicated in the question
the value for phi must be 0 for the velocity equation because the sin of 0 = 0 and the velocity at t=0 is zero
the value for phi for the acceleration must be 0 so that at t=0 we will be taking the cosine of zero which has a value of one so that the acceleration at t=0 will equal its maximum acceleration according to the formula

And by using the following formula to find the value of omega, where k is the spring constant and m is the mass of the block
Subbing in values and simplifying formulas
omega= sqrt(36/4)
omega=3
x(t)=(30)(cos(3t+0)
x(t)=30cos(3t)
v(t)=-(3)(30)(sin(3t+0)
v(t)=-90sin(3t)
a(t)=-(3)^2(30)(cos(3t+0)
a(t)=-270cos(3t)

Formulas Specific to this Question
A graphical representation of these formulas accompanied with an animation
*note: this is a video that represents a dampened spring
Full transcript