Explanation
The correct option is D
Hint: Using trigonometry, convert the given function in suitable form.
Explanation for correct answer:
Step 1: Information required,
The displacement equation is given as,
$$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\cos \omega t$$
Reworking the wave condition as $$\sin $$. Change the $$\cos $$ to $$\sin $$ by $$\cos \omega t$$ = $$ $$ $$\sin \left( {\omega t + \dfrac{\pi}{2}} \right)$$
$$y = \dfrac{1}{{\sqrt a }}\sin \omega t \pm \dfrac{1}{{\sqrt b }}\sin \left( {\omega t + \dfrac{\pi }{2}} \right)………….(1)$$
From the above equation it is observed that the phase difference is $$\dfrac{\pi }{2}$$.
Step 2: Consider general wave equation,
The general wave equation is given as,
$$y = A\sin \left( {\omega t + \theta } \right)…………….(2)$$
Where A represents the amplitude.
Compare amplitudes in both the equations (1) and (2),
Consider amplitudes as $${A_{1,}}{A_2}$$
Then,
$${A_1} = \dfrac{1}{{\sqrt a }}$$ , $${A_2} = \dfrac{1}{{\sqrt b }}$$
Resultant amplitude is given as,
$$A = \sqrt {{A_1}^2 + {A_2}^2} $$
Substituting the $${A_{1,}}{A_2}$$ values in the above equation, we get,
$$A = \sqrt {{{(\frac{1}{{\sqrt a }})}^2} + {{(\frac{1}{{\sqrt b }})}^2}} $$
$$ \Rightarrow A = \sqrt {\dfrac{1}{a} + \dfrac{1}{b}}$$
$$\therefore A = \sqrt {\dfrac{{a + b}}{{ab}}} \\ $$
Therefore, the amplitude of the wave is $$ A = \sqrt {\dfrac{{a + b}}{{ab}}} \\ $$.
$$uy=10sin 60=53√m/s$$
$$⇒t=2uyg=2×53√10=3√s$$
$$Sx=uxt+12axt^2$$
$$1.15=5×t−12a×t^2$$
$$1.15=5×3√−32a$$
$$3a^2=5×1.73−1.15=8.65−1.15$$
$$3a^2=7.5 $$
$$⇒a=153=5m/s^2$$
The displacement of a particle ( in meters ) from its mean position is given by the equation $$y = 0.2\left( {{{\cos }^2}\dfrac{{\pi t}}{2} - {{\sin }^2}\dfrac{{\pi t}}{2}} \right)$$. The motion of the above particle is.
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