A wave is represented by the equation `y=10 sin 2pi (100 t-0.02x)+ 10 sin 2pi (100t +0.02 x).` The maximum amplitude and loop length are respectively.

Options

A. 20 units and 30 units
B. 20 units and 25 units
C. 30 units and 20 units
D. 25 units and 20 units

Explanation

To solve the problem, we need to analyze the given wave equation:

\[ y = 10 \sin(2\pi(100t - 0.02x)) + 10 \sin(2\pi(100t + 0.02x)) \]

### Step 1: Combine the Sine Terms
We can use the trigonometric identity for the sum of sine functions:

\[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right) \]

Let \( A = 2\pi(100t - 0.02x) \) and \( B = 2\pi(100t + 0.02x) \).

Calculating \( A + B \) and \( A - B \):

\[ A + B = 2\pi(100t - 0.02x) + 2\pi(100t + 0.02x) = 2\pi(200t) \]

\[ A - B = 2\pi(100t - 0.02x) - 2\pi(100t + 0.02x) = -2\pi(0.04x) \]

Now substituting back into the identity:

\[ y = 10 \sin A + 10 \sin B = 10 \cdot 2 \sin\left(100\pi t\right) \cos\left(-0.02\pi x\right) \]

This simplifies to:

\[ y = 20 \sin(100\pi t) \cos(0.02\pi x) \]

### Step 2: Determine the Maximum Amplitude
The maximum amplitude of the wave is given by the coefficient of the sine function, which is:

\[ \text{Maximum Amplitude} = 20 \]

### Step 3: Determine the Loop Length
The wave function can be expressed in the form:

\[ y = A \sin(\omega t) \cos(kx) \]

From the equation \( k = 0.02 \times 2\pi \), we can find the wavelength \( \lambda \):

\[ k = \frac{2\pi}{\lambda} \implies \lambda = \frac{2\pi}{0.04\pi} = 50 \]

The length of one loop (or one complete wave cycle) is given by:

\[ \text{Loop Length} = \frac{\lambda}{2} = \frac{50}{2} = 25 \]

### Final Answer
Thus, the maximum amplitude and loop length are:

- Maximum Amplitude: **20**
- Loop Length: **25**