A particle is subjected to two simple harmoni | vedclass

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(A) The two simple harmonic motions are given by:$x_1 = A_1 \sin(\omega t) = \sqrt{7} \sin(5t)$$x_2 = A_2 \sin(\omega t + \phi) = 2\sqrt{7} \sin(5t +

Vedclass · Accepted Answer

$175$

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(A) The two simple harmonic motions are given by:$x_1 = A_1 \sin(\omega t) = \sqrt{7} \sin(5t)$$x_2 = A_2 \sin(\omega t + \phi) = 2\sqrt{7} \sin(5t + \frac{\pi}{3})$Here,$A_1 = \sqrt{7} \ cm$,$A_2 = 2\sqrt{7} \ cm$,$\omega = 5 \ rad/s$,and $\phi = \frac{\pi}{3} = 60^\circ$.The resultant amplitude $A$ is given by the formula:$A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2 \cos\phi}$$A = \sqrt{(\sqrt{7})^2 + (2\sqrt{7})^2 + 2(\sqrt{7})(2\sqrt{7}) \cos(60^\circ)}$$A = \sqrt{7 + 28 + 2(14)(0.5)} = \sqrt{35 + 14} = \sqrt{49} = 7 \ cm = 0.07 \ m$.The maximum acceleration $a_{\max}$ is given by:$a_{\max} = A\omega^2$$a_{\max} = 0.07 	imes (5)^2 = 0.07 	imes 25 = 1.75 \ ms^{-2}$.We are given $a_{\max} = x 	imes 10^{-2} \ ms^{-2}$.$1.75 = x 	imes 10^{-2} \implies x = 175$.

$A$ particle is subjected to two simple harmonic motions as:
$x_1 = \sqrt{7} \sin(5t) \ cm$
and $x_2 = 2\sqrt{7} \sin(5t + \frac{\pi}{3}) \ cm$
where $x$ is displacement and $t$ is time in seconds.
The maximum acceleration of the particle is $x \times 10^{-2} \ ms^{-2}$. The value of $x$ is

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$A$ particle is subjected to two simple harmonic motions as:$x_1 = \sqrt{7} \sin(5t) \ cm$and $x_2 = 2\sqrt{7} \sin(5t + \frac{\pi}{3}) \ cm$where $x$ is displacement and $t$ is time in seconds.The maximum acceleration of the particle is $x \times 10^{-2} \ ms^{-2}$. The value of $x$ is