A travelling wave is produced on a long horizontal string by vibrating an end up and down sinusoidal. The amplitude of vibration is 1.0cm and the displacement becomes zero 200 times per second. The linear mass density of the string is `0.10 kg m^(-1)` and it is kept under a tension of 90 N. (a) Find the speed and the wavelength of the wave. (b) Assume that the wave moves in the positive x-direction and at t = 0 the end x= 0 is at its positive extreme position. Write the wave equation. (c ) Find the velocity and acceleration of the particle at x = 50 cm at time t = 10ms.

Explanation

Let's solve the question step by step.

### Given Data:
- Amplitude of vibration, \( A = 1.0 \, \text{cm} = 0.01 \, \text{m} \)
- Frequency of displacement, \( f = 200 \, \text{times/second} = 200 \, \text{Hz} \)
- Linear mass density of the string, \( \mu = 0.10 \, \text{kg/m} \)
- Tension in the string, \( T = 90 \, \text{N} \)

### Part (a): Find the speed and the wavelength of the wave.

1. **Calculate the speed of the wave \( v \)**:
The speed of the wave on a string is given by the formula:
\[ v = \sqrt{\frac{T}{\mu}} \]
Substituting the values:
\[ v = \sqrt{\frac{90 \, \text{N}}{0.10 \, \text{kg/m}}} = \sqrt{900} = 30 \, \text{m/s} \]

2. **Calculate the wavelength \( \lambda \)**:
The relationship between speed, wavelength, and frequency is given by:
\[ v = f \lambda \]
Rearranging gives:
\[ \lambda = \frac{v}{f} \]
Substituting the values:
\[ \lambda = \frac{30 \, \text{m/s}}{100 \, \text{Hz}} = 0.3 \, \text{m} \]

### Part (b): Write the wave equation.

1. **Form of the wave equation**:
Since the wave moves in the positive x-direction and at \( t = 0 \), the end \( x = 0 \) is at its positive extreme position, the wave equation can be expressed as:
\[ y(x, t) = A \cos(\omega t - kx) \]
where:
- \( A \) is the amplitude,
- \( \omega = 2\pi f \) is the angular frequency,
- \( k = \frac{2\pi}{\lambda} \) is the wave number.

2. **Calculate \( \omega \) and \( k \)**:
- \( \omega = 2\pi \times 100 = 200\pi \, \text{rad/s} \)
- \( k = \frac{2\pi}{0.3} \approx 20.94 \, \text{rad/m} \)

3. **Substituting values into the wave equation**:
\[ y(x, t) = 0.01 \cos(200\pi t - 20.94x) \]

### Part (c): Find the velocity and acceleration of the particle at \( x = 50 \, \text{cm} \) at time \( t = 10 \, \text{ms} \).

1. **Convert units**:
- \( x = 50 \, \text{cm} = 0.5 \, \text{m} \)
- \( t = 10 \, \text{ms} = 0.01 \, \text{s} \)

2. **Find the velocity \( v \)**:
The velocity of the particle is given by:
\[ v = \frac{dy}{dt} = -A \omega \sin(\omega t - kx) \]
Substituting the values:
\[ v = -0.01 \times 200\pi \sin(200\pi \times 0.01 - 20.94 \times 0.5) \]
Calculate:
\[ v = -0.01 \times 200\pi \sin(2\pi - 10.47) \approx -0.01 \times 200\pi \sin(-10.47) \]
Using a calculator, we find \( \sin(-10.47) \approx -0.5 \):
\[ v \approx 5.43 \, \text{m/s} \]

3. **Find the acceleration \( a \)**:
The acceleration of the particle is given by:
\[ a = \frac{d^2y}{dt^2} = -A \omega^2 \cos(\omega t - kx) \]
Substituting the values:
\[ a = -0.01 \times (200\pi)^2 \cos(200\pi \times 0.01 - 20.94 \times 0.5) \]
Calculate:
\[ a = -0.01 \times (200\pi)^2 \cos(2\pi - 10.47) \approx -0.01 \times (200\pi)^2 \cos(-10.47) \]
Using a calculator, we find \( \cos(-10.47) \approx 0.5 \):
\[ a \approx -0.01 \times (200\pi)^2 \times 0.5 \approx -2 \times 10^3 \, \text{m/s}^2 \]

### Final Answers:
(a) Speed: \( 30 \, \text{m/s} \), Wavelength: \( 0.3 \, \text{m} \)
(b) Wave equation: \( y(x, t) = 0.01 \cos(200\pi t - 20.94x) \)
(c) Velocity at \( x = 50 \, \text{cm} \) and \( t = 10 \, \text{ms} \): \( 5.43 \, \text{m/s} \), Acceleration: \( 2000 \, \text{m/s}^2 \)