Question

If the kinetic energy of a particle executing S.H.M. is given by \(\mathrm{K}=\mathrm{K}_{0} \cos ^{2} \omega \mathrm{t}\), then the displacement of the particle is given by \(\ldots \ldots\) (A) $\left\\{\mathrm{K}_{0} / \mathrm{m} \omega^{2}\right\\} \sin \omega \mathrm{t}$ (B) $\left\\{\left(2 \mathrm{~K}_{0}\right) /\left(\mathrm{m} \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t$ (C) \(\left\\{2 \omega^{2} / \mathrm{mK}_{0}\right\\} \sin \omega t\) (D) $\left\\{2 \mathrm{~K}_{0} / \mathrm{m} \omega\right\\}^{1 / 2} \sin \omega t$

Short answer

The displacement of the particle is given by \(x = \left\\{\left(2 K_{0}\right) /\left(m \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t\).

Step-by-step solution

Understand Kinetic and Potential Energy in SHM

In Simple Harmonic Motion, the total energy is the sum of kinetic energy (K) and potential energy (U), and it remains constant throughout the motion: \(E = K + U\) Since the particle is in SHM, we can write the potential energy as: \(U = \frac{1}{2} m \omega^2 x^2\) Where \(x\) is the displacement from the equilibrium position, \(m\) is the mass of the particle, and \(\omega\) is the angular frequency. #Step 2: Expressing total energy#

Express Total Energy in terms of K₀

The total energy (E) is constant, so we can express its initial value in terms of the maximum kinetic energy, K₀: \(E = K_0\) #Step 3: Substituting U and K in the Total Energy equation#

Substitute U and K into E equation

Now, substitute the expressions for U and K in the equation for total energy: \(K_0 = K + U\) \(K_0 = K_0 \cos^2(\omega t) + \frac{1}{2} m \omega^2 x^2\) #Step 4: Solving for displacement#

Solve for displacement x

Re-arrange the equation to solve for the displacement x: \(x^2 = \frac{2(K_0 - K_0 \cos^2(\omega t))}{m \omega^2}\) Now, take the square root of both sides to obtain the displacement expression: \(x = \sqrt{\frac{2 K_0}{m \omega^2}(1-\cos^2(\omega t))}\) Since we know that \(\sin^2(\omega t) = 1 - \cos^2(\omega t)\), we can rewrite the expression in terms of sin: \(x = \sqrt{\frac{2 K_0}{m \omega^2}} \sin(\omega t)\) This matches option (B). Therefore, the displacement of the particle is given by: \(x = \left\\{\left(2 K_{0}\right) /\left(m \omega^{2}\right)\right\\}^{1 / 2} \sin \omega t\)