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10.4 The & Spring Oscillator Conceptual Questions 10, 11, 12, 13 – page 314 Problems 40 page 317

How do the physical aspects of the oscillators affect the Period?

Spring Mass Oscillator Simple Pendulum Mass: Mass: As mass increases the Period (increases/decreases) As mass increases the Period (increases/decreases) As mass decreases the Period (increases/decreases) As mass decreases the Period (increases/decreases)

Spring Constant: Length of Pendulum As K increases the Period (increases/decreases) As L increases the Period (increases/decreases) As K decreases the Period (increases/decreases) AsL decreases the Period (increases/decreases)

Original : Original Displacement: As Δx increases the Period (increases/decreases) As Δx increases the Period (increases/decreases) As Δx decreases the Period (increases/decreases) As Δx decreases the Period (increases/decreases)

Gravity: : As g increases the Period (increases/decreases) As g increases the Period (increases/decreases) As g decreases the Period (increases/decreases) As g decreases the Period (increases/decreases)

The Equation for Period is The Equation for Period is

푇 = 2휋√ 푇 = 2휋√

1 Frequency : 푓 = 푇

Conceptual Questions 10, 11, 12, 13: 10. Suppose that a grandfather clock is running slowly. That is, the it takes to complete each cycle is longer than it should be. Should one shorten or lengthen the pendulum to make the clock keep the correct time? Why?

REASONING AND SOLUTION We can deduce from Equation 10.16 and Example 10 that, for small angles, the period, T, of a simple pendulum is given by T  2 L / g where L is the length of the pendulum.

If a grandfather clock is running slowly, then its period is too long, so one needs to decrease the period to make the clock keep the correct time. Since , we, therefore, need to decrease the length of the pendulum.

11. In principle, the of a simple pendulum and an object on an ideal spring can both be used to provide the basic time interval or period used in a clock. Which of the two kinds of clocks becomes more inaccurate when carried to the top of a high mountain? Justify your answer.

REASONING AND SOLUTION From Equations 10.5 and 10.11, we can deduce that the period of the of an ideal spring is given by T  2 m/ k , where m is the mass at the end of the ideal spring and k is the spring constant. We can deduce from Equations 10.5 and 10.16 that, for small angles, the period, T, of a simple pendulum is given by where L is the length of the pendulum.

In principle, the motion of a simple pendulum and an object on an ideal spring can both be used to provide the period of a clock. However, it is clear from the expressions for the period given above that the period of the mass- spring system depends only on the mass and the spring constant, while the period of the pendulum depends on the due to gravity. Therefore, a is likely to become more inaccurate when it is carried to the top of a high mountain where the value of g will be smaller than it is at sea level.

12. Suppose you were kidnapped and held prisoner by invaders in a completely isolated room, with nothing but a watch and a pair of shoes. Explain how you might determine whether this room is on earth or on the moon.

REASONING AND SOLUTION We can deduce from Equations 10.5 and 10.16 that, for small angles, the period, T, of a simple pendulum is given by where L is the length of the pendulum. This can be solved for the 2 2 acceleration due to gravity to yield: g  4 L /T .

If you were held prisoner in a room and had only a watch and a pair of shoes with shoelaces of known length, you could determine whether this room is on earth or on the moon in the following way: You could use one of the shoelaces and one of the shoes to make a pendulum. You could then set the pendulum into oscillation and use the watch to measure the period of the pendulum. The acceleration due to gravity could then be calculated from the expression above. If the value is close to 9.80 m/s2, then it can be concluded that the room is on earth. If the value is close to 1.6 m/s2, then it can be concluded that the room is on the moon. 13. Two people are sitting on playground swings. One person is pulled back four degrees from the vertical and released, while the other is pulled back eight degrees from the vertical and released. If the two swings are started together, will they both come back to the starting points at the same time? Justify your answer.

REASONING AND SOLUTION The playground swing may be treated, to a good approximation, as a simple pendulum. The period of a simple pendulum is given by T  2 L / g . This expression for the period depends only on the length of the pendulum and the acceleration due to gravity; for angles less than 10° the period is independent of the amplitude of the motion. Therefore, if one person is pulled back 4° from the vertical while another person is pulled back 8° from the vertical, they will both have the same period. If they are released simultaneously, they will both come back to the starting points at the same time.

Problems 40:

40. A simple pendulum is made from a 0.65-m-long string and a small ball attached to its free end. The ball is pulled to one side through a small angle and then released from rest. After the ball is released, how much time elapses before it attains its greatest ? (Hint: Highest speed is at the lowest point. Remember?)

A single period (T) is swinging from the highest position on one side (say the left) to the other side (right) and then back. From knowledge of the Law of Conservation of it should be understood that the highest (speed) is obtained at the lowest point when the maximum Gravitational is converted to . This happens half way across a single swing one way (which is ½ of a cycle). If the period is knowns it should be divided by 4.

퐿 2휋 퐿 휋 퐿 휋 .65 푚 푇 = 2휋√ For this question divide this by 4 푇 = √ = √ = √ = .41 seconds 푔 4 푔 2 푔 2 9.8 푚/푠2

Some Old Favorites: The Simple Pendulum 1. Show that the units are consistent for period equation and the frequency equation. 퐿 푇 = 2휋√ 푔

푚 Plug in Units: 푠 = 2휋√ 푚 푠2

1 2 푠 = = √푠 = 푠 √ 1 푠2

2. Show that sin   for very small angles. When do the two values begin to diverge enough that the difference between them is 15%. (Remember to in . I would suggest using a spreadsheet to make this a quick calculation) Angle Sin % Diff (Radians) 0 0 0.05 0.049979 0% 0.1 0.099833 0% 0.15 0.149438 0% 0.2 0.198669 1% 0.25 0.247404 1% 0.3 0.29552 1% 0.35 0.342898 2% 0.4 0.389418 3% 0.45 0.434966 3% 0.5 0.479426 4% 0.55 0.522687 5% 0.6 0.564642 6% 0.65 0.605186 7% 0.7 0.644218 8% 0.75 0.681639 9% 0.8 0.717356 10% 0.85 0.75128 12% 0.9 0.783327 13% 0.95 0.813416 14% 1 0.841471 16%

3. By what factor does the period (T) for a pendulum change when its length is doubled? 퐿 푇 = 2휋√ 푔 Create the Ratio of the Period (T) for the situations;

for T1 Length = L

for T2 Length = 2L

2퐿 2휋√ √2퐿 푇2 푔 1 √2퐿 = = = = √2 so 푇 = √2 푇 푇 퐿 퐿 퐿 2 1 1 2휋√ √ √ 푔 1

4. Suppose you want to make a clock. You would like the pendulum to swing once per second. How long should it be?

T = 1 second g = 9.8 m/s2 L = ? 퐿 푇2푔 (1 푠푒푐)2(9.8 푚/푠2) 푇 = 2휋√ solving for L gives: 퐿 = = = .248 푚 표푟 24.8 푐푚 푔 4 휋2 4 휋2

Give it a try.

5. Suppose you could take the clock pendulum you just designed for question 3 to the Moon. What would its period be?

T = 1 second g = 1.6 m/s2 L = ? 푇2푔 (1 푠푒푐)2(1.6 푚/푠2) 퐿 = = = .04 푚 표푟 4 푐푚 4 휋2 4 휋2

6. DERIVATION – A pendulum is made from a massless string of length L with a mass M attached. If the pendulum is started at an angle of 10 degrees determine an equation for the a) maximum tension in the string b) minimum tension in the spring.

Length of Pendulum = L , Mass of Pendulum = M Starting Angle = 10 degrees.

Determine Maximum Tension in String.

Logic: The maximum Tension will occur when the Pendulum Bob is moving its fastest. This occurs when the most Energy is converted to Kinetic Energy. This occurs at the lowest point.

Using Law of Conservation of Energy determine the Speed at the Lowest point as a function of M, L, the angle and g

Change in GPE = KE: Mg( L – L cos 10 ) = ½ M V2

Solving for V gives: v = √2푔(퐿 − 퐿푐표푠10)

The at the bottom of the swing is the sum of the Weight plus the .

2푀푔(퐿−퐿푐표푠 10) Total Force = W + Fc = 푀푔 + 퐿

This can be simplified a bit: 푴품 + ퟐ푴품(ퟏ − 풄풐풔 ퟏퟎ)

Minimum Tension in the String occurs when the Pendulum Bob is changing direction, the it has zero velocity.

It is not at Equilibrium so be careful here. However along an axis parallel so the string it is in Equilibrium

The Equation is simple T = Wcos10 or Mgcos10

7. Assign some reasonable hypothetical values for Length (L=2 m) and mass ( M =3 kg) and calculate the values of max and min tension for the 10 degree swing.

Maximum

푴품 + ퟐ푴품(ퟏ − 풄풐풔 ퟏퟎ) =

(ퟑ풌품)(ퟗ. ퟖ풎/풔ퟐ) + (ퟑ풌품)(ퟗ. ퟖ풎/풔ퟐ)(ퟏ − 풄풐풔 ퟏퟎ) =29.8 N

The Centripetal force only adds about .4 N

Minimum

T = Wcos10 or Mgcos10 = (3kg)(9.8 m/s2)cos10 = 28.95 N

SPRING MASS OSCILLATORS

1. Show that the units are consistent for Equation for the Period of a Spring Mass Oscillator.

푚 푇 = 2휋√ 퐾

푘푔 푘푔 푘푔 푥 푚 푘푔 × 푚 1 푠 = 2휋√ = = √ = = = √푠2 = 푠 푁/푚 √ 푁 푁 √ 푘푔 × 푚 √ 1 푚 푠2 푠2

2. For a .5 kg mass, what spring constant is necessary to obtain a period of 1 second?

m = .5 kg T = 1 sec K = ? 푚 4 휋2푚 4 휋2(.5 푘푔) 푇 = 2휋√ solve for K , 푘 = = = 19.7 푁/푚 퐾 푇2 (1 푠푒푐)2

3. Two are set in oscillation by pulling with equal at opposite ends of the system and then releasing them. Upon being released the masses begin to oscillate a. If a single mass is considered “the system” is it in equilibrium? No a single mass is constantly accelerating. b. If a single mass is considered “the system” is the constant (conserved)? No because the velocity is constantly changing. c. If the masses and the spring are considered “the system” is it in equilibrium? No, parts are accelerating. d. If the masses and the spring are “the system” is the momentum constant (conserved)? Yes, the center of gravity does not move.