Trigonometric Identity Problem

**aditya** · August 11th 2009, 09:28 PM

Hi, I'm aditya. Nice to meet you. I have a problem with vibration derivation.
I don't know the answer,

How could possibly the trigonometric function of

become the form

Any chance that the former can be the latter form? I've searched and looked thorough and still don't have the answer until now.

The list of what I've been through is
List of trigonometric identities - Wikipedia, the free encyclopedia
Abramowitz and Stegun, p. 80, 4.4.42
Stumbled upon my own hand-held (writing) derivation which was not satisfactory.

I just don't know how to solve this problem. I think it's a mathematics one. So i tried to submit the problem via this math Forum in there is possibility of someone solving "the unsolved".

It was originally a physics problem. The Vibration derivation of This problem is as follows (it is from the book of Singiresu Rao, Mechanical Vibrations).

Thanks. Any Help would be very appreciated

**Grandad** · August 11th 2009, 10:20 PM

Hello aditya

Welcome to Math Help Forum!

Originally Posted by aditya

Hi, I'm aditya. Nice to meet you. I have a problem with vibration derivation.
I don't know the answer,

How could possibly the trigonometric function of

become the form

Any chance that the former can be the latter form? I've searched and looked thorough and still don't have the answer until now.

The list of what I've been through is
List of trigonometric identities - Wikipedia, the free encyclopedia
Abramowitz and Stegun, p. 80, 4.4.42
Stumbled upon my own hand-held (writing) derivation which was not satisfactory.

I just don't know how to solve this problem. I think it's a mathematics one. So i tried to submit the problem via this math Forum in there is possibility of someone solving "the unsolved".

It was originally a physics problem. The Vibration derivation of This problem is as follows (it is from the book of Singiresu Rao, Mechanical Vibrations).

Thanks. Any Help would be very appreciated

The answer is that, yes, we can write $a\sin x +b\cos x$ as $\sqrt{a^2+b^2}\sin(x+\phi)$ , but, no, this is not the same as $a\sin x +b\cos x =\sqrt{a^2+b^2}$ . It is clearly impossible that it should be so, since $\sqrt{a^2+b^2}$ is a constant, whereas $a\sin x +b\cos x$ is a variable function of $x$ .

The first identity is arrived at like this:

Let $a\sin x +b\cos x= r\sin(x+\phi)$ , for some values of $r$ and $\phi$

Then using the addition formula $\sin(A+B)=\sin A\cos B +\cos A\sin B$ ,

$a\sin x +b\cos x= r(\sin x\cos\phi+\cos x\sin\phi)$

Compare coefficients of $\sin x$ and $\cos x$ :

$a = r\cos\phi$ (1)

$b=r\sin\phi$ (2)

Square (1) and (2) and add:

$a^2 +b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2$ , since $\cos^2\phi+\sin^2\phi=1$

$\Rightarrow r = \sqrt{a^2+b^2}$

Divide (2) by (1):

$\tan\phi = \frac{b}{a}$

So $a\sin x +b\cos x= r\sin(x+\phi)$ , where $r = \sqrt{a^2+b^2}$ and $\tan\phi = \frac{b}{a}$

Does that help to clear things up?

Grandad

**aditya** · August 12th 2009, 10:10 PM

Originally Posted by Grandad

Hello aditya

Welcome to Math Help Forum!The answer is that, yes, we can write $a\sin x +b\cos x$ as $\sqrt{a^2+b^2}\sin(x+\phi)$ , but, no, this is not the same as $a\sin x +b\cos x =\sqrt{a^2+b^2}$ . It is clearly impossible that it should be so, since $\sqrt{a^2+b^2}$ is a constant, whereas $a\sin x +b\cos x$ is a variable function of $x$ .

The first identity is arrived at like this:

Let $a\sin x +b\cos x= r\sin(x+\phi)$ , for some values of $r$ and $\phi$

Then using the addition formula $\sin(A+B)=\sin A\cos B +\cos A\sin B$ ,

$a\sin x +b\cos x= r(\sin x\cos\phi+\cos x\sin\phi)$

Compare coefficients of $\sin x$ and $\cos x$ :

$a = r\cos\phi$ (1)

$b=r\sin\phi$ (2)

Square (1) and (2) and add:

$a^2 +b^2=r^2\cos^2\phi+r^2\sin^2\phi=r^2$ , since $\cos^2\phi+\sin^2\phi=1$

$\Rightarrow r = \sqrt{a^2+b^2}$

Divide (2) by (1):

$\tan\phi = \frac{b}{a}$

So $a\sin x +b\cos x= r\sin(x+\phi)$ , where $r = \sqrt{a^2+b^2}$ and $\tan\phi = \frac{b}{a}$

Does that help to clear things up?

Grandad

Sorry took long to reply..

I have thought o f it too. But the derivation from the book
Make me a little Dizzy..
Mechanical Vibration Singiresu Rao..

The Ampitude Response i think is from 6.. and

and the Phase Difference i think is from 7..

The complete derivation of the equation is as follows in the attachments.

**Grandad** · August 13th 2009, 02:27 AM

Hello aditya

Originally Posted by aditya

Sorry took long to reply..

I have thought o f it too. But the derivation from the book
Make me a little Dizzy..
Mechanical Vibration Singiresu Rao..

The Ampitude Response i think is from 6.. and

and the Phase Difference i think is from 7..

The complete derivation of the equation is as follows in the attachments.

There are, sadly, three mistakes in the text that you are trying to understand.

The differential equation representing forced, damped harmonic motion is correct:

$m\ddot{x} + c\dot{x}+kx = F_0\cos \omega t$ (1)

The Particular Integral can then be written as $x(t)=X\cos(\omega t - \phi)$ (2) Correct.

If we differentiate this twice, we get:

$\dot{x} = -X\omega\sin(\omega t - \phi)$

$\ddot{x} = -X\omega^2\cos(\omega t -\phi)$

Substituting these into equation (1), and re-arrange:

$X[(k-m\omega^2)\cos(\omega t - \phi)-c\omega\sin(\omega t - \phi)]=F_0\cos\omega t$ (3) Correct.

We now use the trig addition formulae to expand $\cos(\omega t - \phi)$ and $\sin(\omega t - \phi)$ and then compare the coefficients of $\cos \omega t$ and $\sin \omega t$ on both sides of the equation, in much the same way as in my first posting.

The coefficient of $\cos\omega t$ on the RHS is $F_0$ . This gives:

$X[(k-m\omega^2)\cos\phi +c\omega\sin\phi] = F_0$ (6) Correct.

The coefficient of $\sin\omega t$ on the RHS is $0$ . This gives:

$X[(k-m\omega^2)\sin\phi - c\omega\cos\phi]=0$ (7). There are two errors in the printed text of this equation!

Equations (6) and (7) are now of the form:

$X[a\cos\phi+b\sin\phi]=F_0$

$a\sin\phi-b\cos\phi=0$

where $a = (k-m\omega^2)$ and $b = c\omega$

From which we get:

$\tan\phi =\frac{\sin\phi}{\cos\phi}= \frac{b}{a}$

$\Rightarrow \sin\phi = \frac{b}{\sqrt{a^2+b^2}}, \, \cos\phi= \frac{a}{\sqrt{a^2+b^2}}$

$\Rightarrow X\Big(\frac{a^2}{\sqrt{a^2+b^2}}+\frac{b^2}{\sqrt{ a^2+b^2}}\Big) = F_0$

$\Rightarrow X\Big(\frac{a^2+b^2}{\sqrt{a^2+b^2}}\Big) = F_0$

$\Rightarrow X = \frac{F_0}{\sqrt{a^2+b^2}} = \frac{F_0}{\sqrt{(k-m\omega^2)^2+c^2\omega^2}}$ (8), not as printed. (Can you see the error?)

and $\phi=\tan^{-1}\Big(\frac{c\omega}{k-mw^2}\Big)$ (9) Correct!

This, then, is the Particular Solution to the Differential Equation. You will also need to consider the solution to the Complementary Function $m\ddot{x} + c\dot{x}+kx = 0$ in order to get the General Solution.

Grandad

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