A windy day

As the wind blast lasted 1 second, and the air leaves the gate with a speed of 0.9m/s, then the air ran a distance of 0.9m after leaving the gate. Hence, the air volume is the gate's area times the distance the air ran, which creates a prism with dimensions 0.9 x 1.5 x 2, that is $2.7m^{3}$ , and this volume of air weights ( $1.22 \times 2.7$ )kg, which is 3.294kg. The air's kinetic energy variation can be expressed as $\Delta E_{k} = E_{k} - E_{k}i = \frac{m_{air}(0.9)^{2}}{2} - \frac{m_{air}1^{2}}{2} \Rightarrow \lvert \Delta E_{k} \rvert = \frac{0.19m_{air}}{2} = \frac{0.19 \times 3.294}{2} = 0.312935 J$

This energy was transferred to the gate, giving it kinetic energy and elastic energy. The maximum distance the gate can go with this energy can be calculated considering the moment in which all the kinetic energy is converted into elastic energy (when the gate will start to go back to its position):

$\lvert \Delta E_{k} \rvert = E_{el} \Rightarrow 0.312935 = \frac{kx^{2}}{2} \Rightarrow 0.312935 = \frac{50x^{2}}{2} \Rightarrow x^{2} = 0.0125172 \Rightarrow x = 0.1118802932m$

Note that "k" is the gate's elastic constant, and "x", the distance the gate moved. As the results are starting to get very big (in number of digits, at least), you can finish the calculations using literal symbols, and you'll reach the same frequency. To calculate the frequency, the time is needed. Then, using Torricelli's Equation, the acceleration can be calculated:

$v^{2} = v_{o}^{2} + 2a\Delta S \Rightarrow x^{2} = 2ax \Rightarrow a = \frac{x^{2}}{2x} \Rightarrow a = \frac{x}{2}$

Note that $x = v$ , because the elastic energy is calculated as $\frac{kx^{2}}{2}$ , and the kinetic energy, as $\frac{m_{air}v^{2}}{2}$ , so, as $k = m$ (numerically), then $x = v$ . Also, note that $v_{o}$ is the gate's initial speed, when it received energy from the air. Now, using the speed's time function, the time can be calculated:

$v = v_{o} +at \Rightarrow x = \frac{x}{2}t \Rightarrow t = 2s$

As there was no energy loss to any object that didn't make part of the system, the gate would, at least hypothetically, keep this vibration forever, because the elastic energy and the kinetic energy would keep this constant convertion. Also, this time that was found (2s) is the time it takes the gate to make just half of the movement. The other half takes two more seconds. Hence, the motion's period is 4s, so its frequency is $\boxed{0.25Hz}$ .