Simple Harmonic Simple Harmonic Motion:Motion:

Abdalla, AymenAbdalla, Aymen Spring 2002Spring 2002 SC.441.L.H, Sec (8773) Instructor: Dr. Roman SC.441.L.H, Sec (8773) Instructor: Dr. Roman

KezerashviliKezerashvili May – 17 - 2002May – 17 - 2002

Introduction:Introduction:

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

ObjectivesObjectives

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

1.1. Two springs.Two springs.

2.2. Triple-beam balance.Triple-beam balance.

3.3. Photo gate accessory.Photo gate accessory.

4.4. Science workshop Interface box.Science workshop Interface box.

5.5. Set of masses.Set of masses.

Equipments:Equipments:

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

Theory:

F = -kxF = -kx (1)(1)

K is the force constant of the spring.K is the force constant of the spring.

From Newton’s second law :From Newton’s second law :

a = F/ma = F/m (2)(2)   Combine (1) and (2)Combine (1) and (2)   a = -kx/m or a= d^2x/dt^2 = -Kx/ma = -kx/m or a= d^2x/dt^2 = -Kx/m (3)(3)

But a varies with x, that: But a varies with x, that: 

^2 = k/m or ^2 = k/m or = = (k/m) (k/m)(4)(4)

  From equation (3)From equation (3)

A = -A = -^2 x or d^2x/dt^2 = -^2 x or d^2x/dt^2 = -^2x^2x(5)(5)

  And from the differential equation : And from the differential equation :

x = Acos(x = Acos(t + t + )) (6)(6)

And from the differential equation : And from the differential equation :

x = Acos(x = Acos(t + t + )) (6)(6)

A is the maximum displacement A is the maximum displacement and it is called amplitude, and it is called amplitude, = = /2/2 is the frequency of vibration, is the frequency of vibration, is is the angular frequency. And (the angular frequency. And (t + t + ) is called the phase of the simple ) is called the phase of the simple harmonic motion, harmonic motion, is called the is called the phase constant.phase constant.

For x from A to –A and back to A is For x from A to –A and back to A is called a cycle ant T is the time for one called a cycle ant T is the time for one complete oscillation (cycle). That:complete oscillation (cycle). That:

cos(cos(t + t + + 2 + 2) = cos() = cos(t + t + )) (7)(7)

So that:So that:

(t + T) + (t + T) + = = t + t + + 2 + 2 or or t = 2t = 2(8)(8)

So:So: T = 2T = 2// (9)(9)

Sub Sub from (4) to (9): from (4) to (9): T= 2T= 2 m/k m/k

(10)(10)

This is true when the mass of the spring ms This is true when the mass of the spring ms is much less than the mass m, which is is much less than the mass m, which is suspended from the spring. And for a suspended from the spring. And for a spring of finite mass ms is:spring of finite mass ms is:

T = 2T = 2 (m + m (m + mss/3)/k/3)/k(11)(11)

From (11) determine k as:From (11) determine k as:

k = 4k = 4^2(m + m^2(m + mss/3)/T^2/3)/T^2 (12)(12)

If we have two springs in series with the If we have two springs in series with the force constant kforce constant k11, k, k22 and x is the sum of the and x is the sum of the displacement of each spring, that is:displacement of each spring, that is:

x = xx = x11 + x + x22

(13)(13)

From Hook’s law:From Hook’s law:

x = Fx = F11/k/k11, x, x22 = F = F22/k/k22, and x =F/k, and x =F/k (14)(14)

Sub these values in (14):Sub these values in (14):

F/k = FF/k = F11/k/k11 + F + F22/k/k22 (15)(15)

Because the springs in series Because the springs in series F = FF = F11 = F = F22, that:, that:

1/k = 1/k1/k = 1/k11 + 1/k + 1/k22 (16)(16)

or: or: k = kk = k11kk22/(k/(k11+k+k22)) (17)(17)

So that T is:So that T is:

T = 2T = 2m(1/km(1/k11 + 1/k + 1/k22)) (18)(18)

Procedure:Procedure: In this experiment we measure the period of In this experiment we measure the period of

oscillation, T for a spring with mass ms and mass oscillation, T for a spring with mass ms and mass m for the object. Then use (12) to determine k, m for the object. Then use (12) to determine k, and repeat the same for a series of two springs.and repeat the same for a series of two springs.

Use the science workshop to measure Frequency Use the science workshop to measure Frequency and Number of cycles.and Number of cycles.

1. Measure the mass of the spring.1. Measure the mass of the spring.2. Suspend the mass of the object larger than the 2. Suspend the mass of the object larger than the

mass of the spring.mass of the spring.3. Start record T period and frequency.3. Start record T period and frequency.4. Increase the mass m and repeat step 4 for total 4. Increase the mass m and repeat step 4 for total

5-7 trails.5-7 trails.5. Connect two springs in series and suspend mass 5. Connect two springs in series and suspend mass

m larger than the mass of the two springs, and m larger than the mass of the two springs, and repeat steps 5 & 5.repeat steps 5 & 5.

Mass of the first single spring = 0.1805Mass of the first single spring = 0.1805

Total suspended mass Total suspended mass m,kgm,kg

Period T,sPeriod T,s Frequency f, Frequency f, HzHz

Square Square

of period of period

T^2, s^2T^2, s^2

K from K from equation equation (12), N/m(12), N/m

.2.2 1.0751.075 .931.931 1.161.16 8.868.86

.25.25 1.191.19 .841.841 1.421.42 8.648.64

.3.3 1.281.28 .781.781 1.641.64 8.678.67

.35.35 1.3631.363 .734.734 1.861.86 8.78.7

.4.4 1.441.44 .694.694 2.072.07 8.88.8

m versus T^2

y = 0.2209x - 0.06

0

0.1

0.2

0.3

0.4

0.5

0 0.5 1 1.5 2 2.5

T^2, s^2

m, K

g

T^2, T^2, s^2s^2

m, m, kkgg

1.161.16 0.20.2

1.421.42 0.250.25

1.641.64 0.30.3

1.861.86 0.350.35

2.072.07 0.40.4

Mean value for k, N/mMean value for k, N/m 30.3630.36

K from the slope of the graphK from the slope of the graph 30.3230.32

% difference% difference .13%.13%

Mass of the second single spring = .035kgMass of the second single spring = .035kg

Total suspended Total suspended mass m,kgmass m,kg

Period Period T,sT,s

Frequency Frequency f, Hzf, Hz

Square Square

of period of period

T^2, s^2T^2, s^2

K from K from equation equation (12), N/m(12), N/m

.2.2 .524.524 1.911.91 .28.28 29.8529.85

.25.25 .583.583 1.7151.715 .34.34 30.530.5

.3.3 .638.638 1.5671.567 .4.4 30.7930.79

.35.35 .687.687 1.4561.456 .47.47 30.530.5

.4.4 .734.734 1.3621.362 .54.54 30.1630.16

m versus T^2

y = 0.768x - 0.0118

0

0.1

0.2

0.3

0.4

0.5

0 0.2 0.4 0.6T^2, s^2

m, k

g

T^2, T^2, ss^̂22

m, m, kgkg

0.280.28 0.20.2

0.340.34 0.250.25

0.40.4 0.30.3

0.470.47 0.350.35

0.540.54 0.40.4Mean value for k, N/mMean value for k, N/m 8.738.73

K from the slope of the graphK from the slope of the graph 8.728.72

% difference% difference .12%.12%

Mass of the series of the two springs = 0.1805Mass of the series of the two springs = 0.1805

Total suspended mass Total suspended mass m,kgm,kg

Period T,sPeriod T,s Frequency f, Frequency f, HzHz

Square Square

of period of period

T^2, s^2T^2, s^2

K from K from equation equation (12), N/m(12), N/m

.3.3 1.4561.456 .687.687 2.122.12 6,926,92

.35.35 1.5521.552 .644.644 2.412.41 6.916.91

.4.4 1.6411.641 .609.609 2.72.7 6.96.9

.42.42 1.6761.676 .597.597 2.812.81 6.916.91

.45.45 1.7251.725 .580.580 33 6.876.87

m versus T^2

y = 0.1714x - 0.0631

0

0.1

0.2

0.3

0.4

0.5

0 1 2 3 4

T^2, s^2

m,

kg

T^2, T^2, s^2s^2 m, kgm, kg

2.122.12 0.30.3

2.412.41 0.350.35

2.72.7 0.40.4

2.812.81 0.420.42

33 0.450.45

Mean value for k, N/mMean value for k, N/m 6.9026.902

K from the slope of the graphK from the slope of the graph 6.86.8

K from (17)K from (17) 6.86.8

% Difference% Difference 1.5%1.5%

Conclusion:Conclusion:

After performing this experiment we can After performing this experiment we can conclude that k for the string is always conclude that k for the string is always constant disregard to the mass of the constant disregard to the mass of the hanging object. hanging object.

In addition, from the three graphs (m In addition, from the three graphs (m versus T^2) for the strings and the series versus T^2) for the strings and the series we can observe the increasing function, we can observe the increasing function, which means that the mass m is directly which means that the mass m is directly proportional to the period of oscillation.proportional to the period of oscillation.

Understanding Problems:Understanding Problems: A mass of 0.2 kg is attached to a spring with a force A mass of 0.2 kg is attached to a spring with a force

constant k equal to 30N/m. if the mass executes constant k equal to 30N/m. if the mass executes simple harmonic motion, what will be its frequency? simple harmonic motion, what will be its frequency?

m = .2kg k = 30N/mm = .2kg k = 30N/m = = /2/2

= = (k/m) (k/m) = = (30/.2) = 12.25(30/.2) = 12.25 = 12.25/2= 12.25/2 = 1.95 Hz = 1.95 Hz Two springs with force constants of spring K1 and K2, Two springs with force constants of spring K1 and K2,

are connected in parallel. What is the spring constant are connected in parallel. What is the spring constant of the combination?of the combination?

F = FF = F11 + F + F22,, F = KxF = Kx

Kx = KKx = K11xx11 + K + K22xx22 x = xx = x11 = x = x22, ,

K = KK = K11 + K + K22