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###### Amplitude derivation help

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1 year ago

Stupid question but.. can anybody help me derive this equation as I seem to be getting stuck with a minus sign instead of a plus.

To find the amplitude A if we are given x_0 and v_0x square Eq. 1 (x_0 = Acos phi) then divide Eq. 2 (v_0x = - omega Asin phi) by omega, square it, and add to the square of Eq. 1. The right side will be A^2 (sin^2 phi + cos^2 phi), which is equal to A^2. The final result is

A = sqrt{x_02 + v_0x^2 / omega^2}

I end up with A = sqrt{x_02 - v_0x^2 / omega^2}

Thanks.

To find the amplitude A if we are given x_0 and v_0x square Eq. 1 (x_0 = Acos phi) then divide Eq. 2 (v_0x = - omega Asin phi) by omega, square it, and add to the square of Eq. 1. The right side will be A^2 (sin^2 phi + cos^2 phi), which is equal to A^2. The final result is

A = sqrt{x_02 + v_0x^2 / omega^2}

I end up with A = sqrt{x_02 - v_0x^2 / omega^2}

Thanks.

Original post by physicsqs

Stupid question but.. can anybody help me derive this equation as I seem to be getting stuck with a minus sign instead of a plus.

To find the amplitude A if we are given x_0 and v_0x square Eq. 1 (x_0 = Acos phi) then divide Eq. 2 (v_0x = - omega Asin phi) by omega, square it, and add to the square of Eq. 1. The right side will be A^2 (sin^2 phi + cos^2 phi), which is equal to A^2. The final result is

A = sqrt{x_02 + v_0x^2 / omega^2}

I end up with A = sqrt{x_02 - v_0x^2 / omega^2}

Thanks.

To find the amplitude A if we are given x_0 and v_0x square Eq. 1 (x_0 = Acos phi) then divide Eq. 2 (v_0x = - omega Asin phi) by omega, square it, and add to the square of Eq. 1. The right side will be A^2 (sin^2 phi + cos^2 phi), which is equal to A^2. The final result is

A = sqrt{x_02 + v_0x^2 / omega^2}

I end up with A = sqrt{x_02 - v_0x^2 / omega^2}

Thanks.

I'm not sure whether I (or others) can even answer the question because of the way the working out here looks due to you writing it on a computer. Could you upload a picture of this working out - the exact same thing you've done here - so I (and others) can have a more clear look at it? I can't guarantee I can work it out myself but I'm sure we'd all appreciate a picture as it looks baffling as it is now.

Reply 2

1 year ago

Original post by moonink

I'm not sure whether I (or others) can even answer the question because of the way the working out here looks due to you writing it on a computer. Could you upload a picture of this working out - the exact same thing you've done here - so I (and others) can have a more clear look at it? I can't guarantee I can work it out myself but I'm sure we'd all appreciate a picture as it looks baffling as it is now.

I found the error now. I didn't square the minus sign in Eq.2 so it becomes positive. Doh. Thanks anyway.

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