A particle is subjected two simple harmonic motions as :
$x_{1} = 7 sin 5 tcm$
and $x_{2} = 27 sin (5 t + \frac{π}{3}) cm$
where x is displacement and $t$ is time in seconds. The maximum acceleration of the particle is $x \times 1 0^{- 2} ms^{- 2}$ . The value of x is :

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TestHub · Accepted Answer

​ x 1 ​ = 7 ​ sin 5 t x 2 ​ = 2 7 ​ sin ( 5 t + 3 π ​ ) ​ From phasor, ∴ Amplitude of resultant SHM = 7 ​ ϕ = tan − 1 7 ​ + 2 7 ​ × 2 1 ​ 2 7 ​ × 3 ​ /2 ​ = tan − 1 2 7 ​ 21 ​ ​ = tan − 1 2 3 ​ ​ ∴ X R ​ = 7 sin ( 5 t + ϕ ) a R ​ = − 7 × 25 sin ( 5 t + ϕ ) ∴ a m a x ​ = 175 cm / sec = 175 × 1 0 − 2 m / sec ​

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