Question

A system containing a ball is oscillating on a frictionless horizontal plane. The position of the mass when its potential energy and its kinetic energy both are equal, is (let $A$ is the amplitude of oscillation)

• A. $A$
• B. $A/\sqrt{2}$
• C. $A/2$
• D. $A/\sqrt{3}$

Answer 1

Answer: (B)

The position is $A/\sqrt{2}$.

Answer 2

Let the position of the ball from the equilibrium position be $x$. In simple harmonic motion (SHM), the potential energy (PE) of the system at position $x$ is given by:

$PE = \frac{1}{2}kx^2$

where $k$ is the effective spring constant of the system.

The kinetic energy (KE) of the ball at position $x$ is given by:

$KE = \frac{1}{2}mv^2$

where $m$ is the mass of the ball and $v$ is its velocity at position $x$.

The velocity $v$ in SHM is related to the amplitude $A$ and position $x$ by:

$v = \omega\sqrt{A^2 - x^2}$

where $\omega$ is the angular frequency of oscillation. For a spring-mass system, $\omega^2 = k/m$. So, the kinetic energy can be written as:

$KE = \frac{1}{2}m(\omega\sqrt{A^2 - x^2})^2 = \frac{1}{2}m\omega^2(A^2 - x^2)$

Substituting $\omega^2 = k/m$, we get:

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

The problem states that the potential energy and kinetic energy are equal:

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

Since $k \neq 0$ (otherwise there would be no oscillation), we can cancel $\frac{1}{2}k$ from both sides:

$x^2 = A^2 - x^2$

Rearranging the equation to solve for $x$:

$x^2 + x^2 = A^2$ $2x^2 = A^2$ $x^2 = \frac{A^2}{2}$

Taking the square root of both sides gives the position $x$:

$x = \pm\sqrt{\frac{A^2}{2}} = \pm\frac{A}{\sqrt{2}}$

The question asks for "the position", and the options are magnitudes. The magnitude of the position where the potential energy equals the kinetic energy is $|x| = \frac{A}{\sqrt{2}}$.