Pergunta

QUESTION THREE (15 MARKS) (a) (i) Give any two examples of systems executing SHM. (ii) Explain two forces acting on a forced harmonic oscillator. (b) A body oscillates in SHM with amplitude 30mm and a frequency of 5. OHz. Calculate the acceleration of the particle at: (i) The extremities of the motion (2 Marks (ii) The centre of the motion (1 Mark) (iii) A position midway between the centre and the extremity (2 Mark (c) An object of mass 0.05 kg is executing SHM. A force of 0.5 N is acting on it when the displacement from the mean position is 0.1 m. (i) Find the time period if the maximum velocity is 5ms^-1 (ii) Calculate the amplitude and maximum acceleration. (iii) Calculate its kinetic and potential energies when the displacement is 0.1 m. Also find the total energy. (2 Marks) (2 Marks) (2 Mar (2 M (2 N
Solução

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RenatoMestre · Tutor por 5 anos
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(a) (i) Two examples of systems executing SHM are:<br />1. A simple pendulum: The pendulum swings back and forth in a periodic motion, which can be approximated as SHM for small angles.<br />2. A mass-spring system: When a mass is attached to a spring and displaced from its equilibrium position, it oscillates back and forth in SHM.<br /><br />(ii) Two forces acting on a forced harmonic oscillator are:<br />1. The restoring force: This force acts in the opposite direction of the displacement and tries to bring the system back to its equilibrium position. It is proportional to the displacement and is given by F = -kx, where k is the spring constant and x is the displacement.<br />2. The applied force: This force is an external force that drives the system to oscillate. It is responsible for maintaining the motion of the system and can be constant or time-varying.<br /><br />(b) For a body oscillating in SHM with amplitude 30mm and a frequency of 50Hz:<br />(i) The acceleration of the particle at the extremities of the motion is given by a = ω^2A, where ω is the angular frequency and A is the amplitude. Substituting the given values, a = (2π * 50)^2 * 0.03 = 15000 m/s^2.<br /><br />(ii) At the centre of the motion, the acceleration of the particle is zero because the displacement is zero.<br /><br />(iii) At a position midway between the centre and the extremity, the acceleration of the particle is given by a = ω^2 * (A/2). Substituting the given values, a = (2π * 50)^2 * 0.015 = 7500 m/s^2.<br /><br />(c) For an object of mass 0.05 kg executing SHM with a force of 0.5 N acting on it when the displacement from the mean position is 0.1 m:<br />(i) The time period can be found using the formula T = 2π√(m/k), where k is the spring constant. Rearranging the formula F = ma for acceleration, we get k = F/m = 0.5/0.05 = 10 N/m. Substituting the values, T = 2π√(0.05/10) = 2π√0.005 = 2π * 0.0707 = 0.445 s.<br /><br />(ii) The amplitude can be found using the formula A = F_max/k, where F_max is the maximum force. Since the force is constant, F_max = 0.5 N. Substituting the values, A = 0.5/10 = 0.05 m. The maximum acceleration is given by a_max = ω^2A, where ω is the angular frequency. Substituting the values, a_max = (2π/0.445)^2 * 0.05 = 355.5 m/s^2.<br /><br />(iii) The kinetic energy (KE) and potential energy (PE) when the displacement is 0.1 m can be calculated using the formulas KE = (1/2)mv^2 and PE = (1/2)kx^2, where v is the velocity and x is the displacement. The velocity can be found using the formula v = ω√(A^2 - x^2). Substituting the values, v = 2π/0.445 * √(0.05^2 - 0.1^2) = 2π/0.445 * √(-0.01) = 2π/0.445 * 0.1 = 1.13 m/s. Substituting the values, KE = (1/2) * 0.05 * (1.13)^2 = 0.0316 J and PE = (1/2) * 10 * (0.1)^2 = 0.05 J. The total energy is the sum of KE and PE, which is 0.0316 + 0.05 = 0.0816 J.
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