Simplifying Laplace Transform of Cosine with Angular Frequency and Phase Shift

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The discussion focuses on finding the Laplace transform of the function f(t) = cos(ωt + φ). The initial approach uses trigonometric identities to express the function in terms of cosine and sine, leading to a standard Laplace transform result. An alternative method attempts to apply the time-shifting property, but it is pointed out that this application is incorrect due to the change in the integral's lower limit. The correct approach involves recognizing that the Laplace transform must account for the Heaviside step function when the starting point of the function is shifted. Overall, the conversation emphasizes the importance of correctly applying properties of the Laplace transform in relation to the function's limits.
dRic2
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Homework Statement


I have to find the L-transform of ##f(x) = cos(\omega t + \phi)##

Homework Equations


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The Attempt at a Solution


The straightforward approach is to write ##cos(\omega t + \phi)## as ##cos(\omega t)cos(\phi) - sin(\omega t)sin(\phi)## and it becomes: $$Lf(s) = \frac {s cos(\phi) - \omega sin(\phi)} {s^2 + \omega ^2}$$.

But can I try this other way ?
##cos ( \omega t + \phi ) = cos \left[ \omega \left( t - \left( - \frac { \phi} { \omega} \right) \right) \right]## and now I can use the t-shift relation to get: $$ Lf(s) = e^{- \left( - \frac {\phi} {\omega} \right) s} L(cos(\omega t)) = e^{ \frac {\phi} {\omega} s} \frac {s} {s^2 + \omega ^2}$$

I don't know if there is a way to simplify my last result or if it is wrong...
 
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You are applying the time-shifting property incorrectly.

Going back to the definition, <br /> F(s) = \int_0^\infty e^{-st} \cos(\omega t + \phi)\,dt. If I set u = t + \frac{\phi}{\omega} I get <br /> F(s) = e^{s\phi/\omega}\int_{\phi/\omega}^\infty e^{-su} \cos(\omega u)\,du. The expression on the right hand side is not e^{s\phi/\omega} times the laplace transform of \cos(\omega t) because the lower limit of the integral is no longer zero.

If \phi/\omega &gt; 0 I can make the right hand side into e^{s\phi/\omega} times the laplace transform of something by adding 0 = e^{s\phi/\omega}\int_0^{\phi/\omega} 0e^{-su}\,du to both sides, which leaves me with the laplace transform of g(t) = \begin{cases} 0 &amp; 0 \leq t &lt; \frac{\phi}{\omega} \\<br /> \cos(\omega t) &amp; t \geq \frac{\phi}{\omega}\end{cases}
 
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Oh, I see, Thank you! In fact now that I pay attention to it, in the book I'm reading the author takes care of the lower limit of the integral not being zero by writing the t-shift like this:
$$L(f(t-t_0)u(t-t_0)) (s) = e^{-t_0s} L(f)(s)$$
where ##u(t)## is the Heaviside step function. Of course here I can not use this property because I can not shift the "starting point" of my function with such a "trick".
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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