A particle makes simple harmonic motion in a straight line 4 cm long. Its velocity while passing through the centre of the line is `16 cm s^(-1)`. The period of the oscillation is

Options

A. `(pi)/(2)s`
B. `(pi)/(4)s`
C. `(pi)/(6)s`
D. `(pi)/(8)s`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Problem
The particle is undergoing simple harmonic motion (SHM) with a total length of 4 cm. The velocity at the center (equilibrium position) is given as 16 cm/s. We need to find the period of the oscillation.

### Step 2: Determine the Amplitude
The total length of the motion is 4 cm, which means the particle moves from one extreme position to the other. The amplitude \( A \) is half of this total length.

\[ A = \frac{L}{2} = \frac{4 \, \text{cm}}{2} = 2 \, \text{cm} \]

### Step 3: Use the Velocity Formula
In SHM, the velocity \( v \) at any position is given by the formula:

\[ v = \omega \sqrt{A^2 - x^2} \]

Where:
- \( \omega \) is the angular frequency,
- \( A \) is the amplitude,
- \( x \) is the displacement from the equilibrium position.

At the center (equilibrium position), the displacement \( x = 0 \). Thus, the formula simplifies to:

\[ v = \omega A \]

### Step 4: Substitute Known Values
We know that \( v = 16 \, \text{cm/s} \) and \( A = 2 \, \text{cm} \). Substituting these values into the equation gives:

\[ 16 = \omega \cdot 2 \]

### Step 5: Solve for Angular Frequency \( \omega \)
Rearranging the equation to solve for \( \omega \):

\[ \omega = \frac{16}{2} = 8 \, \text{rad/s} \]

### Step 6: Find the Period of Oscillation
The period \( T \) of the oscillation is related to the angular frequency \( \omega \) by the formula:

\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):

\[ T = \frac{2\pi}{8} = \frac{\pi}{4} \, \text{s} \]

### Final Answer
The period of the oscillation is \( \frac{\pi}{4} \, \text{s} \).

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