The displacement of a particle along the $x$ axis is given by $ x = a \sin^2 \omega t $. The motion of the particle corresponds to

Question

The root mean square velocity ($v_{\text{rms}}$) of a gas is given by:

Answer 1

Let's analyze the problem statement to determine the type of motion described by the given displacement function.

The displacement of a particle is given as x = a \sin^2 \omega t.

To examine whether this represents simple harmonic motion (SHM), we should review the characteristics of SHM. For a particle in SHM, the displacement x can be typically expressed as:

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

or

x = A \cos(\omega t + \phi)

where A is amplitude, \omega is angular frequency, and \phi is the phase angle.

In our case, the expression for x is a \sin^2 \omega t. Let's use a trigonometric identity to simplify this:

\sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta)

Applying this identity to \sin^2 \omega t, we get:

x = a \left(\frac{1}{2}\right)(1 - \cos 2\omega t) = \frac{a}{2} - \frac{a}{2} \cos 2\omega t

This decomposition of the motion into a constant term \frac{a}{2} and a cosine term -\frac{a}{2} \cos 2\omega t indicates that the particle does not exhibit pure simple harmonic motion.

In simple harmonic motion, the motion should solely consist of the oscillatory (sinusoidal) term, not a constant term as seen here. Hence, the motion described by this function has an additional constant term.

Therefore, the correct option is that the motion of the particle corresponds to non simple harmonic motion.

To rule out other options related to SHM frequency:

The frequency suggestions such as \omega/\pi, 3\omega/2\pi, or \omega/2\pi would imply that motion could fit into the form x = A \sin(\omega t + \phi), which is not the case here.
Hence, options suggesting simple harmonic motion frequencies are also incorrect.

Answer 2