Y=5 sin `pi/2(100t-2x),` what is time period?

Options

A. 0.04 s
B. 1s
C. 0.06 s
D. 0.02 s

Answer

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

### Step 1: Identify the wave equation format
The standard form of a wave equation is:
\[ Y = A \sin(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number,
- \( t \) is time,
- \( x \) is position.

### Step 2: Rewrite the given equation
The given equation can be rewritten as:
\[ Y = 5 \sin\left(\frac{\pi}{2} \cdot 100t - \frac{\pi}{2} \cdot 2x\right) \]
This simplifies to:
\[ Y = 5 \sin\left(50\pi t - \pi x\right) \]

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

### Step 4: Relate angular frequency to time period
The relationship between angular frequency (\( \omega \)) and time period (\( T \)) is given by:
\[ \omega = \frac{2\pi}{T} \]
We can rearrange this to solve for \( T \):
\[ T = \frac{2\pi}{\omega} \]

### Step 5: Substitute the value of \( \omega \)
Substituting \( \omega = 50\pi \) into the equation for \( T \):
\[ T = \frac{2\pi}{50\pi} \]

### Step 6: Simplify the expression
This simplifies to:
\[ T = \frac{2}{50} = \frac{1}{25} \]

### Step 7: Calculate the time period
Calculating \( \frac{1}{25} \) gives:
\[ T = 0.04 \text{ seconds} \]

### Final Answer
The time period \( T \) of the wave is:
\[ \boxed{0.04 \text{ seconds}} \]