The displacement equation of an oscillator is given by x = 5 sin `(2pit+0.5pi)m`. Then its time period and initial displacement are

Options

A. 0.5 s, 5 m
B. 1 s, 2.5 m
C. 0.5 s, 2.5 m
D. 1 s, 5 m

Explanation

To solve the problem, we need to determine the time period and the initial displacement of the oscillator given the displacement equation:

**Displacement equation:**
\[ x = 5 \sin(2\pi t + 0.5\pi) \]

### Step 1: Identify the angular frequency (ω)

The general form of the displacement equation for simple harmonic motion (SHM) is:
\[ x = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

From the given equation, we can see that:
\[ \omega = 2\pi \, \text{(radians per second)} \]

### Step 2: Calculate the time period (T)

The time period \( T \) of an oscillator is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]

Substituting the value of \( \omega \):
\[ T = \frac{2\pi}{2\pi} = 1 \, \text{second} \]

### Step 3: Find the initial displacement (x₀)

The initial displacement \( x_0 \) can be found by substituting \( t = 0 \) into the displacement equation:
\[ x_0 = 5 \sin(2\pi(0) + 0.5\pi) \]
\[ x_0 = 5 \sin(0.5\pi) \]

Knowing that \( \sin(0.5\pi) = \sin(90^\circ) = 1 \):
\[ x_0 = 5 \times 1 = 5 \, \text{meters} \]

### Conclusion

The time period \( T \) is **1 second** and the initial displacement \( x_0 \) is **5 meters**.

### Summary of Results:
- Time Period (T) = 1 second
- Initial Displacement (x₀) = 5 meters

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