The instanteneous displacement of $x$ of a particle executing simple harmonic motion is given by $x = a_{1} sin ω t + a_{2} cos (ω t + π / 6)$ . The amplitude $A$ of oscillation is given by

\sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos \frac{π}{6}}

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\sqrt{a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos \frac{π}{3}}

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\sqrt{a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} cos \frac{π}{6}}

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\sqrt{a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} cos \frac{π}{3}}

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Solution

The correct option is D $\sqrt{a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} cos \frac{π}{3}}$
$x = a_{1} sin ω t + a_{2} cos (ω t + \frac{π}{6})$

$x = a_{1} sin ω t + a_{2} sin (ω t + \frac{π}{6} + \frac{π}{2})$

or $x = a_{1} sin ω t + a 2 sin (ω t + \frac{2 π}{3})$

$∴ A^{2} = a_{1}^{2} + a_{2}^{2} + 2 a_{1} a_{2} cos \frac{2 π}{3}$

$= a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} cos \frac{π}{3}$

$∴ A = \sqrt{a_{1}^{2} + a_{2}^{2} - 2 a_{1} a_{2} cos \frac{π}{3}}$

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Similar questions

A particle is subjected to two simple harmonic motions given as $x_{1} = A_{1} s i n ω t$ and $x_{2} = A_{2} s i n (ω t + \frac{π}{3})$

Find A. the displacement at t = 0

B. the maximum speed of the particle and

C. the maximum acceleration of the particle

y_{1} = A_{1} s i n (ω t - β_{1}), y_{2} = A_{2} s i n (ω t - β_{2})

x_{1} = A_{1} s i n ω t

x_{2} = A_{2} (ω t + \frac{π}{4})

a_{0}^{2} - a_{1}^{2} + a_{2}^{2} . . . . a_{2 n}^{2}

(a_{1}, b_{1}) a n d (a_{2}, b_{2}) i s (a_{1} - a_{2}) x + (b_{1} - b_{2}) y + c = 0

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