Consider a wave propagating in the negative x-direction whose frequency is `100 Hz`. At `t = 5 s`,the displacement associated with the wave is given by
`y=0.5 cos (0.1 x)`
where `x` and `y` are measured in centimetres and `t` in seconds. Obtain the displacement (as a function of x) at `t = 10 s`. What is the wavelength and velocity associated with the wave?

Explanation

To solve the problem step by step, we will follow the given information and derive the required displacement function, wavelength, and velocity.

### Step 1: Understand the given wave equation
The displacement of the wave at \( t = 5 \, s \) is given by:
\[ y = 0.5 \cos(0.1x) \]
This represents a wave propagating in the negative x-direction.

### Step 2: Write the general wave equation
The general form of a wave propagating in the negative x-direction is:
\[ y(x, t) = A \cos(\omega t + kx + \phi) \]
Where:
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( k \) is the wave number,
- \( \phi \) is the phase constant.

### Step 3: Identify parameters from the given equation
From the equation \( y = 0.5 \cos(0.1x) \):
- Amplitude \( A = 0.5 \, \text{cm} \)
- Wave number \( k = 0.1 \, \text{cm}^{-1} \)

### Step 4: Calculate the angular frequency \( \omega \)
Given the frequency \( f = 100 \, \text{Hz} \):
\[ \omega = 2\pi f = 2\pi \times 100 = 200\pi \, \text{rad/s} \]

### Step 5: Determine the phase constant at \( t = 5 \, s \)
At \( t = 5 \, s \), the phase can be expressed as:
\[ \phi + 5\omega = 0 \quad \Rightarrow \quad \phi = -5\omega = -5 \times 200\pi = -1000\pi \]

### Step 6: Write the wave equation at \( t = 10 \, s \)
Now, substituting \( t = 10 \, s \) into the wave equation:
\[ y(x, 10) = 0.5 \cos(200\pi \cdot 10 + 0.1x - 1000\pi) \]
This simplifies to:
\[ y(x, 10) = 0.5 \cos(2000\pi + 0.1x - 1000\pi) \]
Since \( 2000\pi - 1000\pi = 1000\pi \):
\[ y(x, 10) = 0.5 \cos(0.1x + 1000\pi) \]

### Step 7: Simplify using periodic properties of cosine
The cosine function has a periodicity of \( 2\pi \):
\[ \cos(0.1x + 1000\pi) = \cos(0.1x) \quad \text{(since \( 1000\pi \) is a multiple of \( 2\pi \))} \]
Thus, the displacement at \( t = 10 \, s \) is:
\[ y(x, 10) = 0.5 \cos(0.1x) \]

### Step 8: Calculate the wavelength \( \lambda \)
The wave number \( k \) is related to the wavelength \( \lambda \) by:
\[ k = \frac{2\pi}{\lambda} \quad \Rightarrow \quad \lambda = \frac{2\pi}{k} = \frac{2\pi}{0.1} = 20\pi \, \text{cm} \]

### Step 9: Calculate the velocity \( v \)
The wave velocity \( v \) can be calculated using:
\[ v = \omega/k = \frac{200\pi}{0.1} = 2000\pi \, \text{cm/s} \]

### Final Results
1. Displacement at \( t = 10 \, s \):
\[ y(x, 10) = 0.5 \cos(0.1x) \, \text{cm} \]
2. Wavelength \( \lambda \):
\[ \lambda = 20\pi \, \text{cm} \approx 62.83 \, \text{cm} \]
3. Velocity \( v \):
\[ v = 2000\pi \, \text{cm/s} \approx 6283.19 \, \text{cm/s} \]