The amplitude of simple harmonic motion represented by the displacement equation y(cm) = 4(sin 5 $π$ t + $\sqrt{2}$ cos 5 $π$ t) is :

4 cm

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\sqrt{2}

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\sqrt{3}

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\sqrt{2} + 1) c m

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Solution

The correct option is C 4 $\sqrt{3}$ cm
$y = u (s i n 5 π t + \sqrt{2} c o s 5 π t)$
$= u s i n 5 π z t + 4 \sqrt{2} c o s 5 π t$
$y = A s i n w t + B c o s w t$
amplitude $= \sqrt{A^{2} + B^{2}}$
$= \sqrt{(4)^{2} + (4 \sqrt{2})^{2}}$
$= 4 \sqrt{3}$
Option (c)

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Similar questions

y_{1} = 5 (sin 2 π t + \sqrt{3} cos 2 π t)

y_{2} = 5 (sin 2 π t + \frac{π}{4})

Y_{1} = 0.1 sin (100 π t + \frac{π}{3}) and Y_{2} = 0.1 c o s π t .

t = 0

y_{1} = 5 sin (\begin{matrix} 2 π t + \frac{π}{4} \end{matrix})

y_{2} = 2 \sqrt{2} (sin 2 π t + cos 1 π t)

When two displacements represented by $y_{1} = a s i n (t)$ and $y_{2} = b c o s (ω t)$ are superimposed the motion is :

x = 7 s i n ω t +

c o s^{2} \frac{ω t}{2}

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