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Simple Harmonic Motion: Abdalla, Aymen Abdalla, Aymen Spring 2002 Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili SC.441.L.H, Sec.

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Presentation on theme: "Simple Harmonic Motion: Abdalla, Aymen Abdalla, Aymen Spring 2002 Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili SC.441.L.H, Sec."— Presentation transcript:

1 Simple Harmonic Motion: Abdalla, Aymen Abdalla, Aymen Spring 2002 Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili SC.441.L.H, Sec (8773) Instructor: Dr. Roman Kezerashvili May – May –

2 Introduction: Periodic is any motion that repeats it self in any equal interval of time. A vibrating spring and a simple pendulum exhibit periodic motion. The simple harmonic motion is a special type of periodic motion. Periodic is any motion that repeats it self in any equal interval of time. A vibrating spring and a simple pendulum exhibit periodic motion. The simple harmonic motion is a special type of periodic motion.

3 Objectives To study the simple harmonic motion by investigating the period of oscillation of a spring, and to determine the constant of the spring for one spring and two springs in series.

4 1. Two springs. 2. Triple-beam balance. 3. Photo gate accessory. 4. Science workshop Interface box. 5. Set of masses. Equipments:

5 Consider a mass m, attached to a spring. When the spring in a stretched position, a force F acts on the mass, and x is the distance the mass moves from its equilibrium. This force tends to restore the mass to its original position and it is called the restoring force. Also it is opposite to the displacement x and from hooks law. Consider a mass m, attached to a spring. When the spring in a stretched position, a force F acts on the mass, and x is the distance the mass moves from its equilibrium. This force tends to restore the mass to its original position and it is called the restoring force. Also it is opposite to the displacement x and from hooks law. Theory:

6 F = -kx(1) F = -kx(1) K is the force constant of the spring. K is the force constant of the spring. From Newtons second law : From Newtons second law : a = F/m(2) a = F/m(2) Combine (1) and (2) Combine (1) and (2) a = -kx/m or a= d^2x/dt^2 = -Kx/m(3) a = -kx/m or a= d^2x/dt^2 = -Kx/m(3)

7 But a varies with x, that: But a varies with x, that: ^2 = k/m or = (k/m)(4)^2 = k/m or = (k/m)(4) From equation (3) From equation (3) A = -^2 x or d^2x/dt^2 = -^2x (5) And from the differential equation : x = Acos(t + )(6)

8 And from the differential equation : And from the differential equation : x = Acos(t + )(6) x = Acos(t + )(6) A is the maximum displacement and it is called amplitude, = /2 is the frequency of vibration, is the angular frequency. And (t +) is called the phase of the simple harmonic motion, is called the phase constant. A is the maximum displacement and it is called amplitude, = /2 is the frequency of vibration, is the angular frequency. And (t +) is called the phase of the simple harmonic motion, is called the phase constant.

9 For x from A to –A and back to A is called a cycle ant T is the time for one complete oscillation (cycle). That: cos(t + + 2) = cos(t + )(7) cos(t + + 2) = cos(t + )(7) So that: So that: (t + T) + = t or t = 2(8)(t + T) + = t or t = 2(8) So: So: T = 2/(9) T = 2/(9)

10 Sub from (4) to (9): Sub from (4) to (9): T= 2 m/k(10) T= 2 m/k(10) This is true when the mass of the spring ms is much less than the mass m, which is suspended from the spring. And for a spring of finite mass ms is: This is true when the mass of the spring ms is much less than the mass m, which is suspended from the spring. And for a spring of finite mass ms is: T = 2 (m + m s /3)/k(11) T = 2 (m + m s /3)/k(11) From (11) determine k as: From (11) determine k as: k = 4^2(m + m s /3)/T^2(12) k = 4^2(m + m s /3)/T^2(12)

11 If we have two springs in series with the force constant k 1, k 2 and x is the sum of the displacement of each spring, that is: If we have two springs in series with the force constant k 1, k 2 and x is the sum of the displacement of each spring, that is: x = x 1 + x 2 (13) x = x 1 + x 2 (13) From Hooks law: From Hooks law: x = F 1 /k 1, x 2 = F 2 /k 2, and x =F/k(14) Sub these values in (14): Sub these values in (14): F/k = F 1 /k 1 + F 2 /k 2 (15) F/k = F 1 /k 1 + F 2 /k 2 (15)

12 Because the springs in series Because the springs in series F = F 1 = F 2, that: F = F 1 = F 2, that: 1/k = 1/k 1 + 1/k 2 (16) 1/k = 1/k 1 + 1/k 2 (16) or: k = k 1 k 2 /(k 1 +k 2 )(17) or: k = k 1 k 2 /(k 1 +k 2 )(17) So that T is: So that T is: T = 2m(1/k 1 + 1/k 2 )(18) T = 2m(1/k 1 + 1/k 2 )(18)

13 Procedure: In this experiment we measure the period of oscillation, T for a spring with mass ms and mass m for the object. Then use (12) to determine k, and repeat the same for a series of two springs. In this experiment we measure the period of oscillation, T for a spring with mass ms and mass m for the object. Then use (12) to determine k, and repeat the same for a series of two springs. Use the science workshop to measure Frequency and Number of cycles. Use the science workshop to measure Frequency and Number of cycles. 1. Measure the mass of the spring. 2. Suspend the mass of the object larger than the mass of the spring. 3. Start record T period and frequency. 4. Increase the mass m and repeat step 4 for total 5-7 trails. 5. Connect two springs in series and suspend mass m larger than the mass of the two springs, and repeat steps 5 & 5.

14 Mass of the first single spring = Total suspended mass m,kg Period T,s Frequency f, Hz Square of period T^2, s^2 K from equation (12), N/m

15 T^2, s^2 m, k g Mean value for k, N/m K from the slope of the graph % difference.13%

16 Mass of the second single spring =.035kg Total suspended mass m,kg Period T,s Frequency f, Hz Square of period T^2, s^2 K from equation (12), N/m

17 T^2, s ^ 2 m, kg Mean value for k, N/m 8.73 K from the slope of the graph 8.72 % difference.12%

18 Mass of the series of the two springs = Total suspended mass m,kg Period T,s Frequency f, Hz Square of period T^2, s^2 K from equation (12), N/m ,

19 T^2, s^2 m, kg Mean value for k, N/m K from the slope of the graph 6.8 K from (17) 6.8 % Difference 1.5%

20 Conclusion: After performing this experiment we can conclude that k for the string is always constant disregard to the mass of the hanging object. After performing this experiment we can conclude that k for the string is always constant disregard to the mass of the hanging object. In addition, from the three graphs (m versus T^2) for the strings and the series we can observe the increasing function, which means that the mass m is directly proportional to the period of oscillation. In addition, from the three graphs (m versus T^2) for the strings and the series we can observe the increasing function, which means that the mass m is directly proportional to the period of oscillation.

21 Understanding Problems: A mass of 0.2 kg is attached to a spring with a force constant k equal to 30N/m. if the mass executes simple harmonic motion, what will be its frequency? A mass of 0.2 kg is attached to a spring with a force constant k equal to 30N/m. if the mass executes simple harmonic motion, what will be its frequency? m =.2kg k = 30N/m = /2 = (k/m) = (30/.2) = = (k/m) = (30/.2) = = 12.25/2 = 1.95 Hz = 12.25/2 = 1.95 Hz Two springs with force constants of spring K1 and K2, are connected in parallel. What is the spring constant of the combination? Two springs with force constants of spring K1 and K2, are connected in parallel. What is the spring constant of the combination? F = F 1 + F 2,F = Kx Kx = K 1 x 1 + K 2 x 2 x = x 1 = x 2, K = K 1 + K 2 K = K 1 + K 2


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