$${y_1} = 10\sin \left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10

$${y_2} = 5(\sin 3\pi t + \sqrt 3 \cos 3\pi t)$$

$${y_2} = 10\left( {{1 \over 2}\sin 3\pi t + {{\sqrt 3 } \over 2}\cos 3\pi t} \right)$$

$${y_2} = 10\left( {\cos {\pi \over 3}\sin 3\pi t + \sin {\pi \over 3}\cos 3\pi t} \right)$$

$${y_2} = 10\left( {3\pi t + {\pi \over 3}} \right)$$ $$\Rightarrow$$ Amplitude = 10

So ratio of amplitudes = $${{10} \over {10}}$$ = 1

$${x_2} = 5\sqrt 2 \left( {{1 \over {\sqrt 2 }}\sin 2\pi t + {1 \over {\sqrt 2 }}\cos 2\pi t} \right)\sqrt 2 $$

$$ = 10\sin \left( {2\pi t + {\pi \over 4}} \right)$$

$$\therefore$$ $${{{A_2}} \over {{A_1}}} = {{10} \over 5} = 2$$

x(t) = A sin($$\omega$$t + $$\phi$$)

v(t) = A$$\omega$$ cos ($$\omega$$t + $$\phi$$)

2 = A sin$$\phi$$ ...... (1)

2$$\omega$$ = A$$\omega$$ cos$$\phi$$ ....... (2)

From (1) and (2)

tan$$\phi$$ = 1

$$\phi$$ = 45$$^\circ$$

Putting value of $$\phi$$ in equation (1),

$$2 = A\left\{ {{1 \over {\sqrt 2 }}} \right\}$$

$$A = 2\sqrt 2 $$

$$ \therefore $$ x = 2

Applying work energy theorem :

w_g + w_T = $$\Delta$$K

$$-$$mgl(1 $$-$$ cos60$$^\circ$$) = $${1 \over 2}$$mv² $$-$$ $${1 \over 2}$$mu²

v² = u² $$-$$ 2gl(1 $$-$$ cos60$$^\circ$$)

v² = 9 $$-$$ 2 $$\times$$ 10 $$\times$$ 0.5$$\left( {{1 \over 2}} \right)$$

v² = 4

v = 2 m/s