Simple Harmonics Help (1 Viewer)

OMGITzJustin

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Hey people

I'm stuck on a question from the cambridge 3U textbook on Ex 3F, Q10(a):

A particle is moving in SHM on a horizontal line has amplitude 2m. If its speed passing through the centre O of motion is 15m/s, find v2 as a function of the displacement 'x' to the right of O, and find the veolicty and the accleration of the particle when it is 2/3 metres to the right of O
 

Carrotsticks

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We can safely assume that the centre of motion is the orgin, to simplify calculations. This is because the question didn't specify any exact x coordinates ie: When the particle passes through x=4, v= etc etc.

v^2 = n^2 (A^2 - x^2)

We know A=2, so:

v^2 = n^2 (4 - x^2)

When x=0, v=15:

225 = n^2 (4)

n^2 = 225/4.

Therefore v^2 = 225/4 (4 - x^2)

For velocity, directly substitute in x=2/3.

For acceleration, find d/dx (1/2 v^2) and THEN sub in x=2/3.
 
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OMGITzJustin

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We can safely assume that the centre of motion is the orgin, to simplify calculations. This is because the question didn't specify any exact x coordinates ie: When the particle passes through x=4, v= etc etc.

v^2 = n^2 (A^2 + x^2)

We know A=2, so:

v^2 = n^2 (4 + x^2)

When x=0, v=15:

225 = n^2 (4)

n^2 = 225/4.

Therefore v^2 = 225/4 (4 + x^2)

For velocity, directly substitute in x=2/3.

For acceleration, find d/dx (1/2 v^2) and THEN sub in x=2/3.
Thanks carrotsticks, appreciate it
 

Shadowless

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We can safely assume that the centre of motion is the orgin, to simplify calculations. This is because the question didn't specify any exact x coordinates ie: When the particle passes through x=4, v= etc etc.

v^2 = n^2 (A^2 + x^2)

We know A=2, so:

v^2 = n^2 (4 + x^2)

When x=0, v=15:

225 = n^2 (4)

n^2 = 225/4.

Therefore v^2 = 225/4 (4 + x^2)

For velocity, directly substitute in x=2/3.

For acceleration, find d/dx (1/2 v^2) and THEN sub in x=2/3.
isn't it:

v2 = n2 ( a2 - x2 ) ?
 

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