QJEEM23Pha757: JEE Main 2023 Physics - Class 11 JEEM23Pha757 Solution: At a given point of time the value of displacement of a simple…

Given the general equation for displacement in a simple harmonic oscillator:

$x = A \sin(\omega t + \phi)$

At the given time, we have:

$\omega t + \phi = 30^\circ$

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:

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

Now, substitute the values for $A$ and $x$ :

$200 = \frac{1}{2}k\left(\frac{1600 - 1200}{100 \times 100}\right)$

Simplify the equation:

$400 \times 100 \times 100 = k \times 400$

Solve for the force constant, $k$ :

$k = 10^4 \,\text{Nm}^{-1}$

Given that the force constant is expressed as $k = 1.0 \times 10^x \,\text{Nm}^{-1}$ , comparing the values, we get:

$1.0 \times 10^x = 10^4$

Thus, the value of $x$ is $4$ .