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For a particle executing SHM, the kineti...

For a particle executing `SHM`, the kinetic energy `(K)` is given by `K = K_(0)cos^(2) omegat`. The equation for its displacement is

A

`((k_(0))/(momega^(2)))^(1//2) sin omega t`

B

`((2k_(0))/(m omega^(2)))^(1//2)sin omegat`

C

`((2omega^(2))/(m k_(0)))^(1//2)sin omegat`

D

`((2k_(0))/(m omega))^(1//2)sin omegat`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the equation for the displacement of a particle executing Simple Harmonic Motion (SHM) given that its kinetic energy \( K \) is expressed as \( K = K_0 \cos^2(\omega t) \). ### Step-by-Step Solution: 1. **Assume the Displacement Equation**: We start by assuming the standard form of the displacement in SHM: \[ y(t) = A \sin(\omega t) \] where \( A \) is the amplitude. 2. **Find the Velocity**: To find the velocity, we differentiate the displacement with respect to time \( t \): \[ v(t) = \frac{dy}{dt} = \frac{d}{dt}(A \sin(\omega t)) = A \omega \cos(\omega t) \] 3. **Express Kinetic Energy**: The kinetic energy \( K \) of the particle can be expressed in terms of its velocity: \[ K = \frac{1}{2} m v^2 \] Substituting the expression for \( v(t) \): \[ K = \frac{1}{2} m (A \omega \cos(\omega t))^2 = \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) \] 4. **Compare with Given Kinetic Energy**: We are given that: \[ K = K_0 \cos^2(\omega t) \] By comparing the two expressions for kinetic energy, we have: \[ \frac{1}{2} m A^2 \omega^2 \cos^2(\omega t) = K_0 \cos^2(\omega t) \] This implies: \[ \frac{1}{2} m A^2 \omega^2 = K_0 \] 5. **Solve for Amplitude \( A \)**: Rearranging the equation gives: \[ A^2 = \frac{2 K_0}{m \omega^2} \] Taking the square root: \[ A = \sqrt{\frac{2 K_0}{m \omega^2}} \] 6. **Substitute Amplitude into Displacement Equation**: Now substitute the value of \( A \) back into the displacement equation: \[ y(t) = A \sin(\omega t) = \sqrt{\frac{2 K_0}{m \omega^2}} \sin(\omega t) \] 7. **Final Displacement Equation**: Therefore, the equation for the displacement of the particle is: \[ y(t) = \sqrt{\frac{2 K_0}{m \omega^2}} \sin(\omega t) \]
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