A particle on a speing executes shm when the particle is found at x=xmax/2 the speed of the partcle is

A)vx=vmax

B)vx=√3vmax/2

C)vx=√2vmax/2

D)vx=vmax/2

To determine the speed of a particle executing simple harmonic motion (SHM) when it is at a position of \( x = \frac{x_{\text{max}}}{2} \), we can use the principles of SHM and the relationships between position, velocity, and amplitude. Let's break this down step by step.

Understanding Simple Harmonic Motion

In SHM, the position of a particle can be described by the equation:

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

where:

• A is the amplitude (maximum displacement from the equilibrium position).
• \(\omega\) is the angular frequency.
• \(\phi\) is the phase constant.

Velocity in SHM

The velocity of the particle in SHM can be derived from the position function. The velocity \( v(t) \) is given by the derivative of the position with respect to time:

v(t) = -A \omega \sin(\omega t + \phi)

At maximum speed, which occurs at the equilibrium position (x = 0), the speed is:

v_{\text{max}} = A \omega

Finding Velocity at \( x = \frac{x_{\text{max}}}{2} \)

Now, let's find the velocity when the particle is at \( x = \frac{A}{2} \). We can use the conservation of energy principle in SHM, where the total mechanical energy is constant and is given by:

E = \frac{1}{2} k A^2

At any position \( x \), the total energy can also be expressed as the sum of kinetic energy and potential energy:

E = \frac{1}{2} mv^2 + \frac{1}{2} k x^2

Setting these equal gives us:

\frac{1}{2} k A^2 = \frac{1}{2} mv^2 + \frac{1}{2} k x^2

Substituting Values

Substituting \( x = \frac{A}{2} \) into the equation:

\frac{1}{2} k A^2 = \frac{1}{2} mv^2 + \frac{1}{2} k \left(\frac{A}{2}\right)^2

Now, simplifying the potential energy term:

\frac{1}{2} k A^2 = \frac{1}{2} mv^2 + \frac{1}{2} k \frac{A^2}{4}

Rearranging gives:

\frac{1}{2} mv^2 = \frac{1}{2} k A^2 - \frac{1}{8} k A^2

Combining the terms on the right side:

\frac{1}{2} mv^2 = \frac{4}{8} k A^2 - \frac{1}{8} k A^2 = \frac{3}{8} k A^2

Now, solving for \( v^2 \):

v^2 = \frac{3k}{4m} A^2

Since \( \frac{k}{m} = \omega^2 \), we can substitute:

v^2 = \frac{3}{4} A^2 \omega^2

Taking the square root gives:

v = \sqrt{\frac{3}{4}} A \omega = \frac{\sqrt{3}}{2} A \omega

Recalling that \( v_{\text{max}} = A \omega \), we find:

v = \frac{\sqrt{3}}{2} v_{\text{max}}

Final Answer

Thus, when the particle is at \( x = \frac{x_{\text{max}}}{2} \), its speed is:

\text{B) } v_x = \frac{\sqrt{3}}{2} v_{\text{max}}

This means that the correct choice is option B. This analysis illustrates how the position of a particle in SHM directly relates to its speed, showcasing the elegant interplay between kinetic and potential energy in oscillatory motion.