Find the amplitude of a particle due to superposition of following S.H.M, along the same line.
$X_{1} = 2 s i n 50 μ t$
$X_{2} = 10 s i n (50 π t + 37^{0})$
$X_{3} = - 4 s i n 50 π t$
$X_{4} = - 12 c o s 50 π t$

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Solution

$X = X_{1} + X_{2} + X_{3} + X_{4} = 2 sin 50 π t + 10 sin (50 π t + 37 °) - 4 sin 50 π t - 12 cos 50 π t = - 2 sin 50 π t + 10 sin (50 π t + 37 °) - 12 cos 50 π t = sin 50 π t (- 2 + \frac{10}{5} \times 4) + cos 50 π t (10 sin 37 °) - 12 cos 50 π t = 6 sin 50 π t + 6 cos 50 π t - 12 cos 50 π t = 6 sin 50 π t - 6 cos 50 π t$
Amplitude $A = \sqrt{6^{2} + 6^{2}} = 6 \sqrt{2}$

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Similar questions

Find the amplitude (in cm) of a particle due to superposition of two S.H.M. along the same line.

$x_{1} = 5 s i n (t - 37^{o})$

$x_{2} = 6 c o s t$

x_{1} = 3 sin ω t

x_{2} = 5 sin (ω t + 37^{\circ})

x_{3} = - 15 cos ω t

x_{1} = 3 s i n ω t, x_{2} = 5 s i n (ω t + 37^{\circ})

x_{3} = - 15 c o s ω t

x -

x_{1} = 5 sin ω t

x_{2} = 5 sin (ω t + 53^{\circ})

x_{3} = - 10 cos ω t

x_{1} = 10 s i n (ω t + 30^{\circ}

x_{2}

ω t + 60^{\circ}

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