A wave travelling along a strong is described by
`y(x,t)=0.005 sin (80.0x-3.0t)`
in which the numerical constants are in SI units `(0.005m, 80.0 rad m^(-1)` and `3.0 rad s^( -1))`. Calculate (a) the amplitude. (b) the wavelength (c) the period and frequency of the wave. Also , calculate the displacement y of the wave at a distance `x=30.0` cm and time t=20 s?

Explanation

To solve the problem step by step, we will analyze the wave equation given and extract the necessary information to find the amplitude, wavelength, period, frequency, and displacement at specific values of \(x\) and \(t\).

### Given Wave Equation:
\[ y(x,t) = 0.005 \sin(80.0x - 3.0t) \]

### Step 1: Calculate the Amplitude
The amplitude \(A\) of a wave is the coefficient in front of the sine function in the wave equation.

**Solution:**
- From the wave equation, the amplitude \(A\) is:
\[ A = 0.005 \, \text{m} \]

### Step 2: Calculate the Wavelength
The wave number \(k\) is given as \(80.0 \, \text{rad/m}\). The wavelength \(\lambda\) can be calculated using the formula:
\[ \lambda = \frac{2\pi}{k} \]

**Solution:**
- Substituting the value of \(k\):
\[ \lambda = \frac{2\pi}{80.0} = \frac{\pi}{40.0} \, \text{m} \]

### Step 3: Calculate the Period and Frequency
The angular frequency \(\omega\) is given as \(3.0 \, \text{rad/s}\). The frequency \(f\) can be calculated using the relationship:
\[ \omega = 2\pi f \implies f = \frac{\omega}{2\pi} \]

**Solution:**
- Substituting the value of \(\omega\):
\[ f = \frac{3.0}{2\pi} \approx 0.477 \, \text{Hz} \]

The period \(T\) is the reciprocal of frequency:
\[ T = \frac{1}{f} \]

**Solution:**
- Calculating the period:
\[ T = \frac{1}{0.477} \approx 2.095 \, \text{s} \]

### Step 4: Calculate the Displacement at \(x = 30.0 \, \text{cm}\) and \(t = 20 \, \text{s}\)
First, convert \(x\) from centimeters to meters:
\[ x = 30.0 \, \text{cm} = 0.3 \, \text{m} \]

Now substitute \(x\) and \(t\) into the wave equation:
\[ y(0.3, 20) = 0.005 \sin(80.0 \cdot 0.3 - 3.0 \cdot 20) \]

**Solution:**
- Calculate the argument of the sine function:
\[ y(0.3, 20) = 0.005 \sin(24 - 60) = 0.005 \sin(-36) \]
- Using a calculator to find \(\sin(-36^\circ)\):
\[ y(0.3, 20) \approx 0.005 \cdot (-0.5878) \approx -0.002939 \, \text{m} \approx -2.94 \, \text{mm} \]

### Final Answers:
(a) Amplitude: \(0.005 \, \text{m}\)
(b) Wavelength: \(\frac{\pi}{40.0} \, \text{m}\)
(c) Period: \(2.095 \, \text{s}\), Frequency: \(0.477 \, \text{Hz}\)
(d) Displacement at \(x = 30.0 \, \text{cm}\) and \(t = 20 \, \text{s}\): \(-2.94 \, \text{mm}\)