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(C) The given equation is of the form $x = a \sin(\omega t) + b \cos(\omega t)$.This can be rewritten as $x = A \sin(\omega t + \phi)$,where the ampli

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$5$

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(C) The given equation is of the form $x = a \sin(\omega t) + b \cos(\omega t)$.This can be rewritten as $x = A \sin(\omega t + \phi)$,where the amplitude $A$ is given by $A = \sqrt{a^2 + b^2}$.Comparing the given equation $x = 3 \sin(5\pi t) + 4 \cos(5\pi t)$ with the standard form,we have $a = 3$ and $b = 4$.Therefore,the amplitude $A = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 	ext{ cm}$.Thus,the amplitude of the particle is $5 	ext{ cm}$.

If the displacement of a particle is given by $x = 3 \sin (5\pi t) + 4 \cos (5\pi t) \text{ cm}$,what is the amplitude of the particle?

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