1. Given data

• Amplitude: $A = 40\,\text{cm} = 0.4\,\text{m}$
• Displacement at the instant: $y = A\cos 30^\circ$
• Kinetic energy at that instant: $K = 200\,\text{J}$

2. Find the displacement

Since $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, $y = A\frac{\sqrt{3}}{2}$

So, $y^2 = A^2\cdot \frac{3}{4}$

3. Use energy relation in SHM

For a simple harmonic oscillator,

• Total energy: $E = \frac{1}{2}kA^2$
• Potential energy at displacement $y$: $U = \frac{1}{2}ky^2$
• Kinetic energy: $K = E-U = \frac{1}{2}k(A^2-y^2)$

Substitute $y^2 = \frac{3}{4}A^2$: $K = \frac{1}{2}k\left(A^2-\frac{3}{4}A^2\right)$ $K = \frac{1}{2}k\left(\frac{1}{4}A^2\right)$ $K = \frac{1}{8}kA^2$

4. Substitute numerical values

Given $K=200$ J and $A=0.4$ m, $200 = \frac{1}{8}k(0.4)^2$ $200 = \frac{1}{8}k(0.16)$ $200 = 0.02k$

Hence, $k = \frac{200}{0.02} = 10000\,\text{N m}^{-1} = 1.0\times 10^4\,\text{N m}^{-1}$

Therefore, $x = 4$

5. Comparison with stored answer

Stored correct answer = $4$.

This matches the derived answer.