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[Validity]

2005 hsc q5c ii.

how do you find the centre of motion?

 

[Timothy6340]

2005 hsc q5c ii.

how do you find the centre of motion?

by the centre of the motion it means the equibrium so since x=5+2sin(3t-pi/6) the centre of the motion is x=5

 

[Fakeuser]

In any scenario, to find the centre of motion consider the maximum and minimum value of cos and sin which are 1 and -1.

As a result, the amplitude (max value of the function) is 5+2.

The minimum value is 5-2.

Therefore the particle moves between x = 3 and x = 7.

The centre of the motion is the average of these two values and so it is x = 5.

 

[Timothy6340]

In any scenario, to find the centre of motion consider the maximum and minimum value of cos and sin which are 1 and -1.

As a result, the amplitude (max value of the function) is 5+2.

The minimum value is 5-2.

Therefore the particle moves between x = 3 and x = 7.

The centre of the motion is the average of these two values and so it is x = 5.

u just look at the equation since theres a plus 5 it shifts the curve up to x=5 in which the curve moves and and down so its just x=5