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We know the relation between velocity $v$ and displacement $x$ for a simple harmonic oscillator is:
$v = \omega \sqrt{A^2 - x^2}$
where $A$ is the amplitude and $\omega$ is the angular frequency.

From the given data:

1. When $x_1 = 4\text{ m}$, $v_1 = 3\text{ m/s}$:
   $3 = \omega \sqrt{A^2 - 4^2} \implies 9 = \omega^2(A^2 - 16) \quad \text{--- (1)}$

2. When $x_2 = 3\text{ m}$, $v_2 = 4\text{ m/s}$:
   $4 = \omega \sqrt{A^2 - 3^2} \implies 16 = \omega^2(A^2 - 9) \quad \text{--- (2)}$

Dividing equation (1) by equation (2):
$\frac{9}{16} = \frac{A^2 - 16}{A^2 - 9}$
$9(A^2 - 9) = 16(A^2 - 16)$
$9A^2 - 81 = 16A^2 - 256$
$7A^2 = 175$
$A^2 = 25 \implies A = 5\text{ m}$

Substituting $A^2 = 25$ into equation (1):
$9 = \omega^2(25 - 16) = 9\omega^2 \implies \omega^2 = 1 \implies \omega = 1\text{ rad/s}$

Now, the time period $T$ is given by:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{1} = 2\pi\text{ s} \approx 6.283\text{ s}$

Answer:

• Amplitude, $A = 5\text{ m}$
• Time period, $T = 2\pi\text{ s} \approx 6.28\text{ s}$