At a given point of time the value of displacement of a simple harmonic oscillator is given as $\mathrm{y}=\mathrm{A} \cos \left(30^{\circ}\right)$. If amplitude is $40 \mathrm{~cm}$ and kinetic energy at that time is $200 \mathrm{~J}$, the value of force constant is $1.0 \times 10^{x} ~\mathrm{Nm}^{-1}$. The value of $x$ is ____________.

Given the general equation for displacement in a simple harmonic oscillator: <br/><br/> $x = A \sin(\omega t + \phi)$ <br/><br/> At the given time, we have: <br/><br/> $\omega t + \phi = 30^\circ$ <br/><br/> Given the amplitude $A = 40 \,\text{cm}$ and the displacement $x = 40 \times \frac{\sqrt{3}}{2} \,\text{cm} = 20\sqrt{3} \,\text{cm}$, we can write the kinetic energy, $KE$, as: <br/><br/> $KE = \frac{1}{2}k(A^2 - x^2) = 200$ <br/><br/> Now, substitute the values for $A$ and $x$: <br/><br/> $200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$ <br/><br/> Simplify the equation: <br/><br/> $400 \times 100 \times 100 = k \times 400$ <br/><br/> Solve for the force constant, $k$: <br/><br/> $k = 10^4 \,\text{Nm}^{-1}$ <br/><br/> Given that the force constant is expressed as $k = 1.0 \times 10^x \,\text{Nm}^{-1}$, comparing the values, we get: <br/><br/> $1.0 \times 10^x = 10^4$ <br/><br/> Thus, the value of $x$ is $4$.