In a sinusoidal wave the time required for a particular point to move from equilibrium position to maximum displacement is `0.17s`, then the frequency of wave is:

Options

A. 1.47 Hz
B. 0.36 Hz
C. 0.73 Hz
D. 2.94 Hz

Answer

To solve the problem, we need to find the frequency of the wave given the time taken for a point on the wave to move from the equilibrium position to maximum displacement.

### Step-by-Step Solution:

1. **Understand the Motion**: In a sinusoidal wave, the motion of a particle can be described as simple harmonic motion (SHM). The equilibrium position is the mean position, and the maximum displacement is the extreme position.

2. **Identify the Time Interval**: The problem states that the time required for a particle to move from the equilibrium position to maximum displacement is \(0.17 \, \text{s}\).

3. **Relate Time to Period**: In SHM, the time taken to move from the mean position (equilibrium) to the extreme position (maximum displacement) is one-fourth of the total time period \(T\) of the wave. This can be expressed as:
\[ \frac{T}{4} = 0.17 \, \text{s} \]

4. **Calculate the Period \(T\)**: To find the total period \(T\), we can rearrange the equation:
\[ T = 4 \times 0.17 \, \text{s} = 0.68 \, \text{s} \]

5. **Calculate the Frequency \(f\)**: The frequency \(f\) is the reciprocal of the period \(T\):
\[ f = \frac{1}{T} = \frac{1}{0.68 \, \text{s}} \approx 1.47 \, \text{Hz} \]

### Final Answer:
The frequency of the wave is approximately \(1.47 \, \text{Hz}\).

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