The displacement x(in metres) of a particle performing simple harmonic motion is related to time t(in seconds) as `x=0.05cos(4pit+(pi)/4)` .the frequency of the motion will be

Options

A. 0.5 Hz
B. 1.0 Hz
C. 1.5 Hz
D. 2.0 Hz

Answer

To find the frequency of the motion described by the displacement equation \( x = 0.05 \cos(4\pi t + \frac{\pi}{4}) \), we can follow these steps:

### Step 1: Identify the angular frequency
The given equation is in the form of \( x = A \cos(\omega t + \phi) \), where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

From the equation \( x = 0.05 \cos(4\pi t + \frac{\pi}{4}) \), we can identify that:
- The angular frequency \( \omega = 4\pi \) rad/s.

### Step 2: Calculate the frequency
The frequency \( f \) is related to the angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{4\pi}{2\pi} \]

### Step 3: Simplify the expression
Now, simplify the expression:
\[ f = \frac{4\pi}{2\pi} = 2 \text{ Hz} \]

### Conclusion
Thus, the frequency of the motion is \( f = 2 \text{ Hz} \).

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