1. A horizontal cord 10 m long of mass 150 g . a. Calculate the tension the tension in the cord if the wavelength of a 120 Hz wave on it is to be 60 cm ? b. Calculate the quantity of mass to be hunger to give the frequency in (a) above. 2. A standing wave with 3 antinodes oscillates on a string. The speed of waves on the string is \( 15 \mathrm{~m} / \mathrm{s} \) and the length of the string is 0.5 m . calculate the frequency of this standing wave. (NB THE PIPE IS CLOSED BOTH SIDES) An vuvuzela pipe of 40 cm long is oppened both sides. The speed of sound in air \( 360 \mathrm{~m} / \mathrm{s} \). Calculate the frequesncy of sound produced.

Let's solve each part of your question step by step. ### 1. Tension in the Cord **a. Calculate the tension in the cord.** Given: - Length of the cord, \( L = 10 \, \text{m} \) - Mass of the cord, \( m = 150 \, \text{g} = 0.150 \, \text{kg} \) - Frequency of the wave, \( f = 120 \, \text{Hz} \) - Wavelength, \( \lambda = 60 \, \text{cm} = 0.60 \, \text{m} \) First, we can find the wave speed \( v \) using the relationship: \[ v = f \lambda \] Substituting the values: \[ v = 120 \, \text{Hz} \times 0.60 \, \text{m} = 72 \, \text{m/s} \] Next, we can find the linear mass density \( \mu \) of the cord: \[ \mu = \frac{m}{L} = \frac{0.150 \, \text{kg}}{10 \, \text{m}} = 0.015 \, \text{kg/m} \] Now, we can use the wave speed formula related to tension \( T \) and linear mass density \( \mu \): \[ v = \sqrt{\frac{T}{\mu}} \] Rearranging for \( T \): \[ T = \mu v^2 \] Substituting the values: \[ T = 0.015 \, \text{kg/m} \times (72 \, \text{m/s})^2 \] \[ T = 0.015 \times 5184 = 77.76 \, \text{N} \] **b. Calculate the quantity of mass to be hung to give the frequency in (a) above.** The tension in the cord can also be expressed in terms of the mass \( m_h \) hung from the cord: \[ T = m_h g \] Where \( g \approx 9.81 \, \text{m/s}^2 \). Setting the two expressions for tension equal: \[ m_h g = 77.76 \, \text{N} \] Solving for \( m_h \): \[ m_h = \frac{77.76}{9.81} \approx 7.92 \, \text{kg} \] ### 2. Frequency of the Standing Wave **a. Calculate the frequency of the standing wave.** Given: - Speed of waves on the string, \( v = 15 \, \text{m/s} \) - Length of the string, \( L = 0.5 \, \text{m} \) - Number of antinodes = 3 For a string with \( n \) antinodes, the length of the string is given by: \[ L = \frac{n \lambda}{2} \] Where \( n = 3 \) (3 antinodes). Thus: \[ \lambda = \frac{2L}{n} = \frac{2 \times 0.5}{3} = \frac{1}{3} \, \text{m} \approx 0.333 \, \text{m} \] Now, we can find the frequency \( f \): \[ f = \frac{v}{\lambda} = \frac{15 \, \text{m/s}}{0.333 \, \text{m}} \approx 45 \, \text{Hz} \] **b. Calculate the frequency of sound produced by the vuvuzela pipe.** Given: - Length of the vuvuzela pipe, \( L = 40 \, \text{cm} = 0.4 \, \text{m} \) - Speed of sound in air, \( v = 360 \, \text{m/s} \) For a pipe open at both ends, the fundamental frequency is given by: \[ f = \frac{v}{\lambda} \] Where the wavelength for the fundamental mode is: \[ \lambda = 2L = 2 \times 0.4 \, \text{m} = 0.8 \, \text{m} \] Now, substituting into the frequency formula: \[ f = \frac{360 \, \text{m/s}}{0.8 \, \text{m}} = 450 \, \text{Hz} \] ### Summary of Results