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

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None of these

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Solution

The correct option is A

(a) At t = 0, $x_{1} = A_{1} s i n ω t = 0$

and $x_{2} = A_{2} s i n (ω t + \frac{π}{3})$

$= A_{2} s i n (\frac{π}{3}) = \frac{A_{2} \sqrt{3}}{2}$

Thus, the resultant displacement at t = 0 is

(b) The resultant of the two motions is a simple harmonic motion of the same angular frequency $ω$ . The amplitude of the resultant motion is

The maximum speed is

(c) The maximum acceleration is

$a_{m a x} = A ω^{2} = ω^{2} \sqrt{A_{1}^{2} + A_{2}^{2} + A_{1} A_{2}}$

Suggest Corrections

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

a_{1}, a_{2}, a_{3}, a_{4}

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