A transverse wave of frequency 500 Hz and speed 100 m/s is traveling in the positive x direction on a long string. At time t = 0 s the displacements at `x = 0.0` m and at `x = 0.25` m are `0.0` m and `0.02` m, respectively. The displacement at `x = 0.2` m at `t = 5xx10^(-4) s` is

Options

A. `-0.04` m
B. `-0.02` m
C. `0.04` m
D. `0.02` m

Answer

To solve the problem, we need to find the displacement of the wave at \(x = 0.2 \, m\) and \(t = 5 \times 10^{-4} \, s\). We will follow these steps:

### Step 1: Determine the wavelength (\(\lambda\))
The speed of the wave (\(v\)) is given as \(100 \, m/s\) and the frequency (\(f\)) is \(500 \, Hz\). The wavelength can be calculated using the formula:
\[ \lambda = \frac{v}{f} \]
Substituting the values:
\[ \lambda = \frac{100 \, m/s}{500 \, Hz} = 0.2 \, m \]

### Step 2: Calculate the wave number (\(k\))
The wave number (\(k\)) is given by:
\[ k = \frac{2\pi}{\lambda} \]
Substituting the value of \(\lambda\):
\[ k = \frac{2\pi}{0.2} = 10\pi \, rad/m \]

### Step 3: Calculate the angular frequency (\(\omega\))
The angular frequency (\(\omega\)) is calculated as:
\[ \omega = 2\pi f \]
Substituting the frequency:
\[ \omega = 2\pi \times 500 = 1000\pi \, rad/s \]

### Step 4: Write the wave equation
The general form of the transverse wave equation is:
\[ y(x, t) = A \sin(kx - \omega t) \]
We need to find the amplitude \(A\). We know that at \(x = 0.25 \, m\) and \(t = 0 \, s\), the displacement \(y(0.25, 0) = 0.02 \, m\).

### Step 5: Find the amplitude \(A\)
At \(t = 0\):
\[ y(0.25, 0) = A \sin(k \cdot 0.25) \]
Substituting \(k\):
\[ 0.02 = A \sin(10\pi \cdot 0.25) = A \sin(2.5\pi) \]
Since \(\sin(2.5\pi) = -1\):
\[ 0.02 = -A \implies A = -0.02 \, m \]
However, amplitude is taken as a positive value, so \(A = 0.02 \, m\).

### Step 6: Calculate the displacement at \(x = 0.2 \, m\) and \(t = 5 \times 10^{-4} \, s\)
Substituting \(x = 0.2\) and \(t = 5 \times 10^{-4}\) into the wave equation:
\[ y(0.2, 5 \times 10^{-4}) = 0.02 \sin(10\pi \cdot 0.2 - 1000\pi \cdot 5 \times 10^{-4}) \]
Calculating \(kx\):
\[ kx = 10\pi \cdot 0.2 = 2\pi \]
Calculating \(\omega t\):
\[ \omega t = 1000\pi \cdot 5 \times 10^{-4} = 0.5\pi \]
Now substituting these values:
\[ y(0.2, 5 \times 10^{-4}) = 0.02 \sin(2\pi - 0.5\pi) = 0.02 \sin(1.5\pi) \]
Since \(\sin(1.5\pi) = -1\):
\[ y(0.2, 5 \times 10^{-4}) = 0.02 \cdot (-1) = -0.02 \, m \]

### Final Answer
The displacement at \(x = 0.2 \, m\) at \(t = 5 \times 10^{-4} \, s\) is \(-0.02 \, m\).