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Lesson_5.1b_Printable - Important formulae for problem...

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Important formulae for problem solving in oscillations. The displacement of a simple harmonic motion: y = A sin ( ϖ t + δ ) Velocity of a simple harmonic motion: v = A ϖ cos ( ϖ t + δ ) Acceleration of a simple harmonic motion: a = - A ϖ 2 sin ( ϖ t + δ ) = - ϖ 2 y y max = A (amplitude) v max = A ϖ a max = A ϖ 2 k m ϖ = 2 T π ϖ = = 2 π f 1 f T = 2 m T k π = F = - kx 1
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The potential energy of a stretched spring: U = ½ kx 2 The total energy of an oscillating object: The kinetic energy of an oscillating mass: K = ½ mv 2 E = ½ kA 2 The maximum kinetic energy occurs at the equilibrium position: K max = ½ mv max 2 = ½ kA 2 max k v A A m ϖ = = 2
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A 2.0 kg object is attached to a horizontal spring of force constant k = 5 kN.m -1 . The spring is stretched 10 cm from the equilibrium position and released. Find (a) the period, (b) the frequency, and (c) the amplitude of the motion. (d) What is its maximum speed? (e) What is its maximum acceleration? When does the object first reach the equilibrium position? What is the acceleration at this time ? m = 2kg, A = 0.1 m Example 1 k = 5000 Nm -1 , 2 kg (a) Period T is given by: -1 2 kg 2 5000 Nm π = = 0.126 s 2 m T k π = 3
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A 2.0 kg object is attached to a horizontal spring of force constant k = 5 kN.m -1 . The spring is stretched 10 cm from the equilibrium position and released. Find (a) the period, (b) the frequency, and (c) the amplitude of the motion. (d) What is its maximum speed? (e) What is its maximum acceleration? When does the object first reach the equilibrium position? What is the acceleration at this time ? m = 2kg, 10 cm A = 0.1 m Example 1 k = 5000 Nm -1 , 2 kg (b) 1 f T = 1 0.126 s = = 7.94 Hz (c) Since the spring is stretched 10 cm from the equilibrium position, this is the maximum displacement of the object. A = 0.1 m T = 0.126 s 4
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A 2.0 kg object is attached to a horizontal spring of force constant k = 5 kN.m -1 . The spring is stretched 10 cm from the equilibrium position and released. Find (a) the period, (b) the frequency, and (c) the amplitude of the motion. (d) What is its maximum speed? (e) What is its maximum acceleration? When does the object first reach the equilibrium position? What is the acceleration at this time ? f = 7.94 Hz A = 0.1 m Example 1 10 cm 2 kg (d) The angular frequency ϖ is given by: ϖ = 2 π f = 2 π (7.94) = 49.9 Hz The maximum speed v max is given by: v max = A ϖ = 0.1 m × 49.9 s -1 5.0 ms -1 5
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A 2.0 kg object is attached to a horizontal spring of force constant k = 5 kN.m -1 . The spring is stretched 10 cm from the equilibrium position and released. Find (a) the period, (b) the frequency, and (c) the amplitude of the motion. (d) What is its maximum speed? (e) What is its maximum acceleration? (f) When does the object first reach the equilibrium position? What is the acceleration at this time ? ϖ = 49.9 Hz A = 0.1 m Example 1 10 cm 2 kg (e) The acceleration of a SHM is: a = - ϖ 2 y The acceleration becomes maximum when y = A, the amplitude. a max = ϖ 2 A = 49.9 2 s -2 × 0.1 m = 249 ms -2 6
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A 2.0 kg object is attached to a horizontal spring of force constant k = 5 kN.m -1 . The spring is stretched 10 cm from the equilibrium position and released. Find (a) the period, (b) the frequency, and (c) the amplitude of the motion. (d) What is its maximum speed? (e) What is its maximum acceleration? (f)
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