A wire with mass 40.0 g is stretched so that its ends are tied down at points 80.0 cm apart.The wire vibrates in its fundamental mode with frequency 60.0 Hz and with an amplitude at the antinodes of 0.300 cm.
(a) What is the speed of propagation of transverse wave in the wire?
Compute the tension in the wire.
(c) Find the maximum transverse velocity and acceleration of particles in the wire.

Explanation

Let's solve the problem step by step.

### Given Data:
- Mass of the wire, \( m = 40.0 \, \text{g} = 0.040 \, \text{kg} \)
- Length of the wire, \( L = 80.0 \, \text{cm} = 0.80 \, \text{m} \)
- Frequency, \( f = 60.0 \, \text{Hz} \)
- Amplitude at the antinodes, \( A = 0.300 \, \text{cm} = 0.003 \, \text{m} \)

### (a) Speed of Propagation of Transverse Wave in the Wire

1. **Calculate the wavelength (\( \lambda \))**:
- In the fundamental mode, the length of the wire corresponds to half the wavelength:
\[ L = \frac{\lambda}{2} \implies \lambda = 2L = 2 \times 0.80 \, \text{m} = 1.6 \, \text{m} \]

2. **Calculate the speed of the wave (\( V \))**:
- The speed of the wave is given by the formula:
\[ V = f \times \lambda \]
- Substituting the values:
\[ V = 60.0 \, \text{Hz} \times 1.6 \, \text{m} = 96.0 \, \text{m/s} \]

### (b) Compute the Tension in the Wire

1. **Calculate the mass per unit length (\( \mu \))**:
- The mass per unit length is given by:
\[ \mu = \frac{m}{L} = \frac{0.040 \, \text{kg}}{0.80 \, \text{m}} = 0.050 \, \text{kg/m} \]

2. **Use the wave speed formula to find tension (\( T \))**:
- The relationship between speed, tension, and mass per unit length is:
\[ V = \sqrt{\frac{T}{\mu}} \implies T = V^2 \times \mu \]
- Substituting the values:
\[ T = (96.0 \, \text{m/s})^2 \times 0.050 \, \text{kg/m} = 9216 \times 0.050 = 460.8 \, \text{N} \]

### (c) Find the Maximum Transverse Velocity and Acceleration of Particles in the Wire

1. **Calculate the angular frequency (\( \omega \))**:
- The angular frequency is given by:
\[ \omega = 2\pi f = 2 \times 3.14 \times 60.0 \, \text{Hz} \approx 376.8 \, \text{rad/s} \]

2. **Calculate the maximum transverse velocity (\( v_{max} \))**:
- The maximum transverse velocity is given by:
\[ v_{max} = \omega \times A \]
- Substituting the values:
\[ v_{max} = 376.8 \, \text{rad/s} \times 0.003 \, \text{m} \approx 1.13 \, \text{m/s} \]

3. **Calculate the maximum transverse acceleration (\( a_{max} \))**:
- The maximum transverse acceleration is given by:
\[ a_{max} = \omega^2 \times A \]
- Substituting the values:
\[ a_{max} = (376.8 \, \text{rad/s})^2 \times 0.003 \, \text{m} \approx 432 \, \text{m/s}^2 \]

### Final Answers:
- (a) Speed of propagation of transverse wave: \( 96.0 \, \text{m/s} \)
- (b) Tension in the wire: \( 460.8 \, \text{N} \)
- (c) Maximum transverse velocity: \( 1.13 \, \text{m/s} \) and maximum transverse acceleration: \( 432 \, \text{m/s}^2 \)