A travelling wave on a long stretched string along the positIve x-axis is given by `y = 5mm e^(((t)/(5s) - (x)/(5cm))^2)`. Using this equation answer the following questions.
At t = 0, x = 0, the displacement of the wave is

Options

A. 0
B. `oo`
C. 5mm
D. 10mm

Answer

To find the displacement of the wave at \( t = 0 \) and \( x = 0 \), we can follow these steps:

### Step 1: Write down the wave equation
The wave is described by the equation:
\[ y = 5 \, \text{mm} \, e^{\left(\frac{t}{5 \, \text{s}} - \frac{x}{5 \, \text{cm}}\right)^2} \]

### Step 2: Substitute the values of \( t \) and \( x \)
We need to evaluate the displacement \( y \) at \( t = 0 \) and \( x = 0 \):
\[ y = 5 \, \text{mm} \, e^{\left(\frac{0}{5 \, \text{s}} - \frac{0}{5 \, \text{cm}}\right)^2} \]

### Step 3: Simplify the expression
Calculating the terms inside the exponent:
\[ \frac{0}{5 \, \text{s}} = 0 \quad \text{and} \quad \frac{0}{5 \, \text{cm}} = 0 \]
Thus, we have:
\[ y = 5 \, \text{mm} \, e^{(0 - 0)^2} = 5 \, \text{mm} \, e^{0} \]

### Step 4: Evaluate \( e^{0} \)
Since \( e^{0} = 1 \):
\[ y = 5 \, \text{mm} \times 1 = 5 \, \text{mm} \]

### Conclusion
Therefore, the displacement of the wave at \( t = 0 \) and \( x = 0 \) is:
\[ \boxed{5 \, \text{mm}} \]

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