The displacement of an elastic wave is given by the function $y = 3 \sin \omega t + 4 \cos \omega t$,where $y$ is in $cm$ and $t$ is in $s$. Calculate the resultant amplitude. Also,find the initial phase (epoch).
(A) The given equation is $y = 3 \sin \omega t + 4 \cos \omega t$ ... $(1)$
$A$ standard harmonic wave equation is given by $y = a \sin (\omega t + \phi)$.
Expanding this,we get $y = a \sin \omega t \cos \phi + a \cos \omega t \sin \phi$ ... $(2)$
Comparing equations $(1)$ and $(2)$,we have:
$a \cos \phi = 3$ ... $(3)$
$a \sin \phi = 4$ ... $(4)$
Squaring and adding equations $(3)$ and $(4)$:
$a^2 \cos^2 \phi + a^2 \sin^2 \phi = 3^2 + 4^2$
$a^2 (\cos^2 \phi + \sin^2 \phi) = 9 + 16$
$a^2 = 25$
$a = 5 \ cm$ (Resultant amplitude).
To find the initial phase $\phi$,divide equation $(4)$ by equation $(3)$:
$\frac{a \sin \phi}{a \cos \phi} = \frac{4}{3}$
$\tan \phi = \frac{4}{3}$
$\phi = \tan^{-1} (\frac{4}{3}) \approx 53.13^\circ$.
$A$ standard harmonic wave equation is given by $y = a \sin (\omega t + \phi)$.
Expanding this,we get $y = a \sin \omega t \cos \phi + a \cos \omega t \sin \phi$ ... $(2)$
Comparing equations $(1)$ and $(2)$,we have:
$a \cos \phi = 3$ ... $(3)$
$a \sin \phi = 4$ ... $(4)$
Squaring and adding equations $(3)$ and $(4)$:
$a^2 \cos^2 \phi + a^2 \sin^2 \phi = 3^2 + 4^2$
$a^2 (\cos^2 \phi + \sin^2 \phi) = 9 + 16$
$a^2 = 25$
$a = 5 \ cm$ (Resultant amplitude).
To find the initial phase $\phi$,divide equation $(4)$ by equation $(3)$:
$\frac{a \sin \phi}{a \cos \phi} = \frac{4}{3}$
$\tan \phi = \frac{4}{3}$
$\phi = \tan^{-1} (\frac{4}{3}) \approx 53.13^\circ$.
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