The equation of a simple harmonic wave is given by `Y = 5sin""pi/2(100t - x)`, where x and y are in metre and time is in second. The time period of the wave (m seconds) will be

Options

A. 0.04
B. 0.01
C. 1
D. 5

Explanation

To find the time period of the wave given by the equation \( Y = 5 \sin\left(\frac{\pi}{2}(100t - x)\right) \), we can follow these steps:

### Step 1: Identify the wave equation format
The general form of a simple harmonic wave is given by:
\[ Y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number.

### Step 2: Rewrite the given equation
The given equation is:
\[ Y = 5 \sin\left(\frac{\pi}{2}(100t - x)\right) \]
We can rewrite this as:
\[ Y = 5 \sin\left(50\pi t - \frac{\pi}{2} x\right) \]
This allows us to identify \( \omega \) and \( k \).

### Step 3: Identify the angular frequency
From the rewritten equation, we can see that:
\[ \omega = 50\pi \]

### Step 4: Relate angular frequency to time period
The relationship between angular frequency \( \omega \) and time period \( T \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( \omega \):
\[ 50\pi = \frac{2\pi}{T} \]

### Step 5: Solve for the time period \( T \)
To find \( T \), we can rearrange the equation:
\[ T = \frac{2\pi}{50\pi} \]
This simplifies to:
\[ T = \frac{2}{50} = \frac{1}{25} \text{ seconds} \]

### Step 6: Convert to milliseconds
Since the question asks for the time period in milliseconds, we convert seconds to milliseconds:
\[ T = \frac{1}{25} \times 1000 = 40 \text{ milliseconds} \]

### Final Answer
Thus, the time period of the wave is:
\[ \boxed{40 \text{ ms}} \]

---