The displacement of a particle executing S.H.M. is x = 5 sin (20`pit`). Then its frequency will be

Options

A. `20pi` Hz
B. 20 Hz
C. 10 Hz
D. 10`pi` Hz

Answer

To find the frequency of a particle executing simple harmonic motion (S.H.M.) given its displacement equation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the displacement equation**: The displacement of the particle is given as:
\[ x = 5 \sin(20 \pi t) \]

2. **Compare with the standard form**: The standard form of the displacement in S.H.M. is:
\[ x = A \sin(\omega t + \phi) \]
where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.

3. **Extract parameters**: From the given equation \(x = 5 \sin(20 \pi t)\), we can identify:
- Amplitude \(A = 5\)
- Angular frequency \(\omega = 20 \pi\)
- Phase constant \(\phi = 0\)

4. **Relate angular frequency to frequency**: The relationship between angular frequency \(\omega\) and frequency \(f\) is given by:
\[ \omega = 2 \pi f \]

5. **Substitute the value of \(\omega\)**: We can substitute \(\omega = 20 \pi\) into the equation:
\[ 20 \pi = 2 \pi f \]

6. **Solve for frequency \(f\)**: To find \(f\), we divide both sides by \(2 \pi\):
\[ f = \frac{20 \pi}{2 \pi} = 10 \]

7. **Conclusion**: The frequency of the S.H.M. is:
\[ f = 10 \text{ Hz} \]

### Final Answer:
The frequency of the particle executing S.H.M. is **10 Hz**.

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