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

Given

$$y=A\cos⁡30^{\circ\ }$$

and

A=40 cm=0.4 m

So displacement is

$$y=0.4\cdot\frac{\sqrt{3}}{2}$$

$$y=0.2\sqrt{3}$$

For SHM, total energy is

$$E=\frac{1}{2}kA^2$$

Potential energy at displacement y is

$$U=\frac{1}{2}ky^2$$

Kinetic energy is

$$K=E−U$$

So

$$K=\frac{1}{2}k(A^2-y^2)$$

Given kinetic energy is

200J

thus

$$200=\frac{1}{2}k(A^2-y^2)$$

Now

$$A^2=(0.4)^2=0.16$$

and

$$y^2=(0.2\sqrt{3})^2=0.12$$

So

$$200=\frac{1}{2}k(0.16-0.12)$$

$$200=\frac{1}{2}k(0.04)$$

$$200=0.02k$$

$$k=\frac{200}{0.02}=10000$$

$$k=1.0\times10^4\text{ N/m}$$

Comparing with

$$1.0\times10^x$$

x=4