Physics 2 Assignment - Week 4 Simple Harmonic Motion PDF

Physics 2: Assignment – Week 4: Simple Harmonic Motion Name: ______________________ Nickname: ________________ ID (short): _________ 1. A spring of spring constant 𝑘 is vertically hung from a ceiling. A weight with a mass of 500 [g] is hung at the other end of...

Physics 2: Assignment – Week 4: Simple Harmonic Motion Name: ______________________ Nickname: ________________ ID (short): _________ 1. A spring of spring constant 𝑘 is vertically hung from a ceiling. A weight with a mass of 500 [g] is hung at the other end of the spring, causing it to stretch for 𝑥0 = 156.2 [cm] and stop at the point O. Then the weight is pulled down by a distance 20.0 [cm] and released, causing the weight to oscillate up and down with frequency of 1.00 [Hz]. Let the magnitude of the gravitational acceleration be 𝑔 = 9.81 𝟐𝟎 [cm] [m/s2]. a. Find the spring constant 𝑘. ∑ ⃗⃗⃗𝐹 = 0 −𝑚𝑔 + 𝑘𝑥0 = 0 𝑚𝑔 0.500 × 9.81 𝑘= = = 3.14 𝑘 = _______________ [N/m] Unit ______ 𝑥0 1.562 b. Find the period of oscillation 𝑇 and the angular velocity 𝜔. 1 1 2𝜋 𝑇 = 𝑓 = 1.00 = 1.00 [s] and 𝜔 = = 6.28 [rad/s] 𝑇 𝑇 =__________ Unit ________ and 𝜔 =____________ Unit _________ c. Find the amplitude of this oscillation 𝐴. 0.20 𝐴 =______________________ Unit [m] ___________ d. Express the displacement from the equilibrium position O (downward) as a function of times 𝑦(𝑡). 𝑦(𝑡) = 𝐴 cos 𝜔𝑡 = 0.200 cos(6.28𝑡) e. Find the displacement at time 𝑡 = 100 [s]. 𝑦(𝑡 = 100) = 𝐴 cos 𝜔𝑡 = 0.200 cos(6.28 × 100) 0.190 The displacement = ________________ [m] Unit ______ f. Find the speed when the weight is passing through the position O. The object will get a maximum speed. For any function of 𝑣 = −𝜔𝐴 sin 𝜔𝑡, and 𝑣 = 𝜔𝐴 cos 𝜔𝑡, both will get the speed of 𝑣max = 𝜔𝐴 (because at this maximum sin 𝜔𝑡 and cos 𝜔𝑡 is one) 𝑣 = 𝜔𝐴 = 6.28 × 0.2 = 1.26 [m/s] The speed _________________Unit________ 1 Physics 2: Assignment – Week 4: Simple Harmonic Motion Name: ______________________ Nickname: ________________ ID (short): _________ g. Find the velocity as a function for times 𝑣(𝑡). Looking up the situation types, we get: 𝑣 = −𝜔𝐴 sin 𝜔𝑡 = −1.26 sin 6.28𝑡 h. Find the maximum magnitude of acceleration 𝑎max = 𝜔2 𝐴 = (2𝜋)2 (0.2) = 7.90 [m/s2] 𝑎 = _________________Unit________ i. Find the acceleration as a function of times 𝑎(𝑡) 𝑎(𝑡) = −𝐴𝜔2 cos 𝜔𝑡 = −7.89 cos(6.28𝑡) 2. There is an object with a mass of 2.0 [kg] that undergoes a simple harmonic motion with an amplitude of 2.0 [m]. When the object is at a point P, 1.0 [m] away from the center of vibration in the positive-y direction, the object receives a force of 4.0 [N] in the negative-y direction. Answer the following questions without considering the gravity. a. Find the angular frequency and then the period of the simple harmonic motion of this object. 𝐹 = −𝑚𝜔2 𝑦 −4.0 = −2.0 × 𝜔2 × 1.0 1.4 [rad/s] The angular frequency = _________ Unit _________ 4.4 [s] The period = ___________ Unit ___________ b. What is the speed of this object when the object is at point P? 𝑣 = 𝜔𝐴 cos 𝜃 2.4 [m/s] The speed = _________ Unit _________ c. What is the speed of this object when it passes through the center of vibration? 𝑣max = 𝜔𝐴 2.8 [m/s] The speed = _________ Unit _________ 2 Physics 2: Assignment – Week 4: Simple Harmonic Motion Name: ______________________ Nickname: ________________ ID (short): _________ d. Find the acceleration of this object when it is at point Q which is 0.50 [m] in the negative direction away from the center of vibration. 𝑎 = −𝜔2 𝑦 = −(1.4)2 × (−0.5) 1.0 [m/s] The acceleration = _________ Unit _________ e. What is the maximum value of the acceleration of this object? 𝑎 = 𝐴𝜔2 = 2.0 × (1.4)2 2 4.0 Unit _________ The acceleration = _________ [m/s ] 3