A particle is subjected two simple harmonic motions as : x1 = √7 sin 5t cm

Question

A particle is subjected two simple harmonic motions as :

\(x_1 = \sqrt 7 \ sin5t \ cm\)

and  \(x_2 = 2 \sqrt 7 sin \left(5t + \frac{\pi}{3}\right) cm\)

where x is displacement and t is time in seconds. The maximum acceleration of the particle is

x × 10^–2 ms^–2 . The value of x is :

(1) 175

(2) \(25 \sqrt{7}\)

(3) \(5 \sqrt{7}\)

(4) 125

Answer 1

Correct option is : (1) 175

\(x_1 = \sqrt 7 \ sin5t\)

\(x_2 = 2 \sqrt7 \ sin \left(5t+ \frac{\pi}{3}\right)\)

From phasor,

\(\therefore\)  Amplitude of resultant SHM = 7

\(\phi = tan^{-1} \frac{2 \sqrt{7} \times \sqrt 3/2}{\sqrt 7 + 2 \sqrt7\times \frac{1}{2}} = tan ^{-1} \frac{\sqrt 21}{2 \sqrt{7}} = tan ^{-1} \frac{\sqrt3}{2}\)

\(\therefore \ X_R = 7 \ sin (5 t+ \phi)\)

\(a_R = -7 \times 25 \ sin (5 t+ \phi)\)

\(\therefore \ a _ {max} = 175 \ cm/sec = 175 \times 10^{-2} \ m/sec\)