A particle executes linear simple harmonic motion with an amplitude of $3 \,cm$. When the particle is at $2\,cm$ from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is :

Question

What is the fundamental frequency of an open organ pipe with length $ L $ and speed of sound $ v $?

Answer 1

To solve the problem of finding the time period of the particle executing simple harmonic motion (SHM), we need to start by considering the basic equations of motion for SHM.

For a particle executing SHM, the displacement x from the mean position at any time t can be given as:

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

Where:

A is the amplitude.
\omega is the angular frequency, related to the time period T by \omega = \frac{2\pi}{T}.
\phi is the phase constant.

Given in the problem:

Amplitude A = 3 \, \text{cm}
Displacement x = 2 \, \text{cm}

When the particle's velocity is equal to its acceleration in magnitude, we use the following formulas:

1. Velocity of the particle, v = \pm \omega \sqrt{A^2 - x^2}

2. Acceleration, a = -\omega^2 x

Given that the magnitudes are equal:

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

Substitute the given values:

\sqrt{3^2 - 2^2} = \omega x

\sqrt{9 - 4} = \omega \cdot 2

\sqrt{5} = 2\omega

Solving for \omega:

\omega = \frac{\sqrt{5}}{2}

Now, we find the time period T:

T = \frac{2\pi}{\omega} = \frac{2\pi}{\frac{\sqrt{5}}{2}}

T = \frac{4\pi}{\sqrt{5}}

Thus, the time period of the particle is \frac{4\pi}{\sqrt{5}} \, \text{seconds}. This matches with the correct option given.

Answer 2