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 solve this problem, we need to use the basic equations of simple harmonic motion (SHM). The velocity \(v\) of a particle in SHM at a distance \(x\) from the mean position is given by: 

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

where \(A\) is the amplitude and \(\omega\) is the angular frequency.

First, we compute the angular frequency \(\omega\) using the speed of the particle at the mean position. At the mean position, \(x = 0\), and the velocity \(v_{max} = 10 \, \text{m/s}\) is the maximum. Thus:

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

Substitute the given values:

\(10 = \omega \times 4\)

Solve for \(\omega\):

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

Next, we find the distance \(x\) when the velocity \(v = 5 \, \text{m/s}\):

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

Simplifying this gives:

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

Divide both sides by 2.5:

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

Square both sides to remove the square root:

\(4 = 16 - x^2\)

Rearrange to solve for \(x^2\):

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

Hence, solving for \(x\):

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

Therefore, the distance of the particle from the mean position where the velocity becomes 5 m/s is \(2\sqrt{3} \; m\), matching the correct option.