Question 368: Oscillations

Question

The potential energy of a particle of mass 0.1 kg moving along the $X$-axis, is given by $U=5 x(x-4) \mathrm{J}$, where $x$ is in metres. Choose the wrong option.

Accepted Answer

The potential energy $ U $ of a particle with mass 0.1 kg moving along the $ X $-axis is described by:
$ U = 5x(x-4) \, 	ext{J} $
where $ x $ is the position in meters.
The equation expands to:
$    U = 5x^2 - 20x    $
To find the force $ F $, we take the derivative of $ U $ with respect to $ x $:
$    \frac{dU}{dx} = 10x - 20    $
The force is given by the negative derivative of the potential energy:
$    F = -\frac{dU}{dx} = -10x + 20    $
Assuming a shifted position $ x = (X + 2) $, the force becomes:
$    F = -10X    $
The force is zero when $ X = 0 $ or equivalently, $ x = 2 \, 	ext{m} $. Therefore, $ x = 2 \, 	ext{m} $ is the equilibrium or mean position about which the particle performs simple harmonic motion (SHM).
The spring constant $ k $ is 10, leading to a period of oscillation $ T $ calculated as follows:
$    T = 2\pi \sqrt{\frac{m}{k}} = 2\pi \sqrt{\frac{0.1}{10}} = \frac{\pi}{5} \, 	ext{s}    $
At the mean position, the kinetic energy of the particle is at a maximum, which is consistent with SHM behavior, where the force acting on the particle varies.

Answer

The speed of the particle is maximum at $x=2 \mathrm{~m}$

Answer

The particle executes simple harmonic motion

Answer

The period of oscillation of the particle is $\frac{\pi}{5} \mathrm{~s}$

Answer

None of the above

Oscillations

The potential energy of a particle of mass 0.1 kg moving along the X-axis, is given by U=5 x(x-4) \mathrm{J}, where x is in metres. Choose the wrong option.

Choose the correct answer: