Question

A particle performing simple harmonic motion is such that its amplitude is 4 m and speed of particle at mean position is 10 m/s. Find the distance of particle from mean position where velocity becomes 5 m/s.

• A. \(\sqrt3\; m\)
• B. \(2\sqrt3\; m\)
• C. \(\frac{\sqrt3}{2}\; m\)
• D. \(\frac{1}{\sqrt2}\; m\)

Answer 1

The Correct Option is B

To determine the solution, we will utilize the fundamental equations governing simple harmonic motion (SHM). The velocity \(v\) of an object undergoing SHM, at a displacement \(x\) from its equilibrium position, is defined as:

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

In this formula, \(A\) represents the amplitude, and \(\omega\) denotes the angular frequency.

Initially, we calculate the angular frequency \(\omega\) by considering the particle's speed at its equilibrium position. At the equilibrium position, \(x = 0\), and the velocity reaches its maximum value, \(v_{max} = 10 \, \text{m/s}\). This leads to the equation:

\(v_{max} = \omega A\)

Substituting the provided values yields:

\(10 = \omega \times 4\)

Solving for \(\omega\):

\(\omega = \frac{10}{4} = 2.5 \, \text{rad/s}\)

Subsequently, we determine the displacement \(x\) when the velocity is \(v = 5 \, \text{m/s}\):

\(5 = 2.5 \sqrt{4^2 - x^2}\)

Simplifying this equation gives:

\(5 = 2.5 \sqrt{16 - x^2}\)

Dividing both sides by 2.5:

\(2 = \sqrt{16 - x^2}\)

To eliminate the square root, we square both sides:

\(4 = 16 - x^2\)

Rearranging the equation to solve for \(x^2\):

\(x^2 = 16 - 4 = 12\)

Finally, we solve for \(x\):

\(x = \sqrt{12} = 2\sqrt{3} \, \text{m}\)

Therefore, the particle's distance from the mean position when its velocity is 5 m/s is \(2\sqrt{3} \; m\), which corresponds to the correct answer.