# What does a simple pendulum swing?

### Pendulum and Pendulum Laws

##### A body that can swing freely back and forth about an axis that does not go through the center of gravity is a pendulum.

A distinction is made between 1. simple or mathematical and 2. compound pendulums.

The former, which exist only in the imagination, are a heavy point suspended from a weightless thread; E.g. a small gold or platinum ball on a silk thread would come very close to this. All others, including the clock pendulums, are physical or composite pendulums.

The movement of a pendulum from the highest to the opposite highest point is called a half-oscillation, and a pendulum movement from one highest point to the other and back to the first is called an oscillation. The point of a pendulum, which on its own would swing just as fast as the whole pendulum, is called the center of oscillation and its distance from the bend point of the pendulum spring is the mathematical pendulum length of the physical pendulum.

Since the calculation of a physical pendulum cannot be carried out without many and cumbersome mathematical calculations, only a simple formula for the mathematical pendulum, from which the required can be easily derived, should find space here:

T = l: g

##### (In plain language: pi times the root l over g, cannot be displayed correctly in HTML)

It means T the duration of a simple half oscillation (impact duration), pi the known value 3.14159 ..., l the mathematical pendulum length explained above and g the acceleration due to gravity, which for middle latitudes is about 9.81 m / sec^{2} is assumed, and, since the earth has no spherical shape, but is flattened at the poles, between the values 9.8322 m / sec^{2} (for the poles) and 9.7803 m / sec^{2} (on the equator) fluctuates. From the latter values, the formula above gives the number 991.02 mm for the mathematical length of a second pendulum on the equator. For Paris, g = 9.810 m / sec^{2}, l therefore 993.93 mm; for Berlin g 0 9.813 m / sec^{2}, l thus 994.25 mm and for the North Pole g = 9.832 mm / sec^{2} and l = 996.45 mm. This information applies to an altitude of sea level.

A standard seconds pendulum clock that goes to the 0^{th} Latitude goes right, it would take almost 4 minutes in a day to go to a pole!

##### The following explanations and examples apply to middle latitudes, unless otherwise noted.- The period of oscillation is independent of the mass and material of the vibrating body (a wooden ball and a lead ball on threads of the same length vibrate at the same speed)
- With a small deflection angle, the duration of an oscillation is almost independent of the size of the angle within certain limits: Principle of isochronism - ie equality of times and means that a pendulum can, under certain circumstances, make an arc of 1.5 ° and an arc of 2.5 ° exactly can go through the same time (endeavor of watchmaking)
- The duration of the pendulum oscillations is proportional to the square root of the pendulum lengths (a quarter-meter pendulum needs half as much time to oscillate as a one-meter pendulum

T_{1}: T

_{2}= Root l

_{1}: Root l

_{2}

##### The oscillation numbers of two pendulums behave in reverse as the square roots of the pendulum lengths

Example: Two pendulums are 1 m and 0.25 m in length, where the root 1 = 1 and the root 0.25 = 0.5. The quarter-meter pendulum makes twice as many oscillations in the same time as the one-meter pendulum.

n_{1}: n

_{2}= Root l

_{2}: Root l

_{1}

##### The duration of a pendulum oscillation is inversely proportional to the square root of the acceleration due to the gravitational pull (the greater the latter - towards the poles - the shorter the duration of the oscillation, i.e. the faster the pendulum oscillates).

### To determine the mathematical length of a pendulum from the given number of oscillations per minute

##### A pendulum-seconds pendulum makes 60 semi-oscillations per minute and is 994 mm long. If the number of beats per minute of another pendulum is denoted by n_{m}, then according to the above rule 4 one has by inversion and squaring for the mathematical pendulum length sought:

l: l_{x}= n

^{2}

_{m}/ n

^{2}or l

_{x}= n

^{2}/ n

^{2}m * l

##### For n (number of blows) = 60 and l = 994 mm, this is calculated as follows:

l_{x}= (60 / n

_{m})

^{2}* 994 = 3600 * 994 / n

^{2}

_{m}= 3578400 / n

^{2}

_{m}(in mm), or, rounded, l

_{x}= 3580000 / n

^{2}

_{m}(Length in mm).

##### Example: The clock in the Dresden Zwinger (Math. Phys. Salon) in regulator form takes 5/4 seconds to half an oscillation. How big is the mathematical pendulum length?

Given: n_{m}= 60 / (5/

_{4}) = 60 * 4/5 = 48. Wanted: l =?

Resolution: 3580000/48

^{2}= 3580000 / 2304 =

**1553.82 mm.**

### Calculation of the length difference of a physical pendulum for a determined path difference

##### The solution to this problem is based on the formula

delta l = l * (2.25 * delta t) / t##### Here delta t (delta, Greek first letter of difference = difference) is the difference in length of the pendulum to be eliminated, l is the pendulum length in mm found in the previous section, 2.25 is a value found in practice, delta t is that in 24 hours = 1 day observed time difference in minutes, and t the observation time given in minutes.

If you insert the value found for l (in mm) after the previous section, then is

delta l = delta t * (2.25 * 3580000) / (24 * 60) = delta t * 5593.75 / n^{2}

_{m}or rounded

delta l = delta t * 5600 / n

^{2}

_{m}(delta t in minutes per day, n pendulum strokes per minute)

##### Example: A pendulum seconds clock has a time difference of one second in a day. How much does the pendulum length have to be changed?

Given: delta t = 1 sec. = 1/60 min, n = 60. Wanted: delta l =?Resolution: delta l = delta t * 5600 / n

^{2}

_{m}= (1 * 5600) / ( 60 * 60 * 60 ) = 56 / (60 * 36) = 14 / 540 = 7 / 270 =

**0.02592 mm**

### Calculation of the rotation of the regulating screw nut of a pendulum for an observed path difference

##### The work of regulating a pendulum clock can be carried out much faster if one determines the pitch of the pendulum thread, which is done by counting the number of threads for a certain number of millimeters that match the threads and dividing the number of millimeters by the number of threads, from this the displacement of the pendulum lens is calculated for one division of the graduation attached to the regulating nut and divides the change in length required according to the previous section by this value. If you turn the regulating screw nut to the correct side, the path difference is eliminated apart from a small remainder.

Task: A clock with a seconds pendulum has a time difference of exactly 100 seconds in 4 days. 30 turns of the pendulum thread are 24 mm long, and the regulating screw nut is divided into 8 parts. How many tick marks does it have to be rotated?

Given: delta t = 100 sec. In 4 days = 25 sec. In 1 day = 25/60 min in 1 day. Number of pendulum strokes n_{m}= 60, number of tick marks = 8. Wanted: The required number of tick marks =?

Resolution: There is a length shift of 24 / (30 * 8) = 0.1 mm per division.

According to section 2, delta l = (25 * 5600) / (60 * 3600) = (25 * 56) / (60 * 36) = (5 * 14) / (12 * 9) = 70/108 = 0.648

The number of graduation marks you are looking for = 0.648 / 0.1 =

**6.5 lines**

### Calculation of the additional weights of a pendulum

##### The pendulum of observation clocks, normal clocks and the like, from which very precise rate results must be required, must never be stopped unnecessarily, because otherwise some time, about 2-3 days, would pass before the clock has completely resumed the previous rate. Likewise, clocks with heavy and long pendulums should also be handled carefully so that the lamellae (which are often the very thin springs of the pendulum suspension) are not bent and when they are straightened, tension is created on one side.

To prevent the pendulum from stopping, a small table is attached to the pendulum rod, about 40 mm square or (if round) 50 mm in diameter, the upper surface of which should be half the mathematical pendulum length of the pendulum in question from the bending point of the pendulum spring and that serves to accommodate the so-called additional weights. The latter consist of narrow strips of aluminum sheet about 0.3-0.4 mm thick, about 3-4 mm wide, and their weight is calculated according to the formula

delta p = (8 * delta t * P) / t ##### calculated, in which 8 is a value that depends on the mass distribution of the pendulum (Clemens Riefler gives the value 7.7 for his pendulum weighing 7 kg), delta t is the observed time difference in seconds, P is the total pendulum weight in grams and t indicates the observation time in seconds. The numbers for delta t and t should be given in seconds, because after the clock has been pre-regulated, the remaining error is very small. The following additional weights should be available:

For a daily time difference of 1 second 2 pieces

For a daily rate difference of 0.5 seconds, 2 pieces

For a daily rate difference of 0.2 seconds, 4 pieces

For a daily rate difference of 0.1 seconds, 4 pieces

For a daily path difference of 0.05 seconds, 2 pieces

For a daily path difference of 1 second, the additional weight would be 8/86400 = 1/10800 or, according to Riefler, 7.7 / 86400 = 7/11220, i.e. on average 1/11000 of the pendulum weight. It should not be forgotten that the pendulum should carry an additional weight for one second when pre-regulating, so that the additional weight can be reduced under certain circumstances when the clock is moving forward, i.e. when it is supposed to slow down.**If you add an additional weight, the watch advances according to the additional weight** , since the center of gravity of the pendulum moves upwards, the effective pendulum length is therefore shorter. Even with very simple watches, the use of additional weights has proven itself and is more convenient and more advantageous for the watch than screwing around on the regulating screw nut.

Task: How heavy must the additional weights of a second pendulum of 6 kg be? (you determine the weight for 1 second per day and then the other weights)

Given: delta t = 1 sec., P = 6 kg = 6000 g, t = 1 day = 86400 sec. Wanted: delta p =?Resolution: delta p = (8 * delta t * P) / t = (8 * 1 * 6000) / 86400 = (8 * 60) / 864 = 60/108 = 5/9 =

**0.555 g**

##### You need 2 pieces of 0.555 g each for 1 second

You need 2 pieces of 0.277 g each for 0.5 sec

4 pieces of 0.111 g each are required for 0.2 sec

You need 4 pieces of 0.055 g each for 0.1 sec

You need 2 pieces of 0.027 g each for 0.05 sec

### Calculation of the change in length of the body with changing temperature

##### The watchmaker knows from the behavior of the pendulums and the compensation disturbances that all bodies, with very few exceptions, expand in the warmth, i.e. become longer and contract in the cold.

Therefore, in order to be able to calculate the size of this change, the expansion coefficients obtained through precise experiments (these are numbers that are multiplied by the length of the body, by how much the length of the body changes by 1 ° Celsius when heated or cooled) .

If the difference in length is denoted by delta l, the number of degrees Celsius by which the body is heated by t and the coefficient of expansion by alpha, so is

delta l = l * alpha * t### Expansion coefficients for changes in length (between 0 ° and 100 ° C)

material | alpha for 1 ° C | material | alpha for 1 ° C |

aluminum | 0,0000 238 | platinum | 0,0000 090 |

antimony | 0,0000 108 | silver | 0,0000 197 |

lead | 0,0000 292 | Tantalum | 0,0000 065 |

bronze | 0,0000 175 | bismuth | 0,0000 134 |

Iron, flux iron | 0,0000 120 | tungsten | 0,0000 043 |

Iron, mild steel | 0,0000 117 | zinc | 0,0000297 |

Iron, cast iron | 0,0000 104 | tin | 0,0000 230 |

Iron, welding iron | 0,0000 122 | Thermometer glass | 0,0000 081 |

Invar (64% iron, 36% nickel) | 0,0000 010 | Jena Glass No. 59 | 0,0000 059 |

Hard rubber | 0,0000 770 | ||

Constantan | 0,0000 152 | porcelain | 0,0000 030 |

copper | 0,0000 165 | quartz | 0,0000 004 |

Magnalium | 0,0000 240 | Wood, maple | 0,0000 050 |

Brass | 0,0000 184 | Wood, oak | 0,0000 062 |

Nickel silver | 0,0000 180 | Wood, fir tree | 0,0000 035 |

nickel | 0,0000 135 |

##### Task: A zinc rod is 85 cm long at + 10 ° C. How much does it expand when heated to 48 ° C?

Given: l = 85 cm = 850 mm, alpha = 0.0000 297, delta t = 48 ° - 10 ° = 38 ° C. Wanted: delta l =?Resolution: delta l = l * alpha * t = 850 * 0.0000297 * 38 =

**0.959 mm**

### Calculation of the length of the body with changing degrees of heat

##### In the last section it was shown how the size of the change in the length of the body is found, and the body length is now to be determined under the same conditions, i.e. in the previous section only the more or less in length and now the changed overall length.

This can be found by adding the change in length to the original length or subtracting it when it cools down. The current length is denoted by l_{2}, the former length with l_{1}, the change as before with delta l, then delta l = l_{1} * alpha * t.

Hence: l_{2}= l

_{1}± l

_{1}* alpha * t = l

_{1}* (1 ± alpha * t)

##### The lower sign (-) must be taken with decreasing temperature.

To ensure that no disagreements can arise when measuring tools and workpieces, especially limit gauges, plug gauges, ring gauges, snap gauges, gauge blocks, are delivered, a reference temperature of 20 ° C has been set for them by DIN.

Also for the material from which these objects are made, it is determined that the change in length at 1 m length and 1 ° C temperature difference may only be 0.0115 mm. The measuring authorities only certify the values for 20 ° C. The expansion coefficient should therefore only be max.alpha = 0.0000115 for these materials.

Task: The zinc rod of a grate pendulum is 600 mm long. What is the length of this when it is brought from 12 ° C to 25 ° C?

Given: l

_{1}= 600 mm, alpha = 0.0000297, t = 25 ° C - 12 ° C = 13 ° C. Wanted: l

_{2}= ?

Resolution: l

_{2}= l

_{1}* (1 + alpha * t) = 600 * (1 + 0.0000297 * 13) = 600 * (1 + 0.0003861) = 600 * 1.0003861 = 6 * 100.03861 =

**600.23166 mm**

### Rate difference in pendulum clocks due to changing degrees of heat

##### While section 2 deals with the length change of a pendulum necessary to eliminate a path difference, the opposite is now to be done, namely the time difference is calculated that arises when the pendulum rod changes its original length as a result of different degrees of heat. In the mentioned section is the formula

delta l = (l * 2.25 * delta t) / t##### specified. So it is

delta l * t = l * 2.25 * delta t##### from which one obtains:

delta t = (delta l * t) / (2.25 * l)##### According to Section 1, l = 3580000 / n^{2}_{m} and after section 5 delta l = l * alpha * t.

Task: A tower clock has a 2½ seconds pendulum whose pendulum rod is made of iron. The temperature changes from + 25 ° C to -15 ° C. How does the clock go in winter when it has gone correctly in summer?

Given: n_{m}= 60 / 2½ = 24, alpha = 0.0000120, t = 40 ° C. Wanted: delta t =?

Resolution: First you determine the pendulum length:

l = 3580000/24

^{2}= 3580000/576 = 6215 mm.

The change in length:

delta l = l * alpha * t = 6215 * 0.0000120 * 40 = 2.9832 mm.

The time difference:

delta t = (2.9832 * 86400) / (2.25 * 6215) = 25774848/1398375 =

**~ 18.4 sec.**

##### The clock thus advances approximately 18.4 seconds in one day in winter.

### Automatic compensation of the clock rate differences in the event of temperature and air pressure fluctuations (compensation)

##### In Sections 5 to 7 it has been shown what influence the temperature changes can have on the clocks, and now it is to be indicated how the errors arising as a result can be eliminated or at least reduced.

If a pendulum rod expands as a result of greater heat, the pendulum becomes longer and the clock slows down. If you now attach a second rod vertically upwards at the lower end of the pendulum, which, held at the bottom, has to expand upwards, and connects with this the pendulum lens, which can have any shape, you can easily see that the rod lengths and the Materials can be selected in such a way that the pendulum lens neither rises nor falls at different temperatures, so the clock continues evenly, ie runs correctly.

The resulting pendulum shapes are called- Rust pendulum
- Mercury pendulum
- Quartz pendulum
- Nickel steel compensation pendulum

The former, modeled on a grate, usually consist of 5 bars, 3 of which are made of the same material, usually iron or steel, while the 2nd and 4th are made of copper, brass or zinc to compensate for heat.

The length of the copper, brass or zinc rods is to be chosen so that the change in length is equal to the sum of the changes in the three iron or steel rods growing downwards. If the value for alpha does not agree with reality, the length would have to be changed, which is done simply by switching on another piece of the compensation rod by means of a row of holes and a pin.

This shape, which was often and almost exclusively used in the past, also in precision clocks, is rarely used today, because free expansion is hindered if it is properly guided and, in the other case, the pendulum loses its internal strength.

The lower end of the mercury pendulum has one or two vessels made of glass, iron or steel filled with mercury. The large amount of mercury only slowly takes on the heat that the thin pendulum rod, which is surrounded by air, has much earlier, and it is not really suitable for shipping.

The easily breakable quartz pendulum, the rod of which is made of fused quartz, suffers from the same inconvenience.

The nickel steel pendulums consist of Invar rods which, due to their small size, only require short compensation pieces made of iron, brass, steel or aluminum, mostly assembled. The pendulum body is supported in the center of gravity, so it can expand in any direction without changing gear.

The warm air goes up in every room, including in the watch case, so that there are layers of air of different warmth that must have a disadvantageous effect on the compensation. An attempt is made to compensate for this error by covering the inside of the housing with sheet copper, which, as a very good conductor of heat, brings the cold from the bottom up and the heat from the top down.

The changing air pressure exerts a further disruptive influence on the rate of the clocks. This produces path differences of 0.012-0.018 seconds, depending on the shape of the pendulum lens, in one day with a 1 mm change in the barometer level. This error can be corrected by placing the entire clock under an airtight glass bell, which is pumped out to a barometer reading of around 650 mm, or, if it should always be accessible for any other reason, receiving air pressure compensation (this is an aneroid barometer with applied weight, attached to about ¼ of the mathematical pendulum length from above on the pendulum rod). The clocks placed under a bell jar are regulated by changing the air pressure.

Task: A nickel steel Invar pendulum of 994 mm length is to be fitted with a compensation tube made of brass. How long does this have to be?

Given: l_{1}= 994 mm, alpha

_{1}= 0.000001, alpha

_{2}= 0.0000184. Wanted: l

_{2}= ?

Resolution: l

_{1}* alpha

_{1}= l

_{2}* alpha

_{2}; 994 * 0.000001 = l

_{2}* 0,0000184 =

l

_{2}= (994 * 0,000001) / 0,0000184 = (994 * 10) / 184 =

**54 mm**

### Regulating a pendulum clock for another location

##### It is not uncommon for the manufacturer of the precision watch to demand that the watch should work correctly not at the place of manufacture but at the place of the customer.

In section 1 it is stated that the value g is subject to great changes, which depend on the shape of the earth, so that even a pendulum clock cannot work correctly everywhere if it is brought to another location without change.

The resulting time difference can be calculated with a fair amount of accuracy using the simple formula

delta t = 44 * (g_{1}- g

_{2}) sec.

##### The only thing to note is that the two values for g are to be taken in centimeters. The following small table should make it easier to solve such tasks.

The values are based on the sea surface.

Geographical width | Gravity acceleration in cm / sec ^{2} | Geographical width | Gravity acceleration in cm / sec ^{2} |

0° | 978,030 | 50° | 981,07 |

10° | 978,25 | 51° | 981,16 |

20° | 978,69 | 52° | 981,24 |

30° | 979,36 | 53° | 981,33 |

40° | 980,17 | 54° | 981,42 |

41° | 980,26 | 55° | 981,50 |

42° | 980,35 | 56° | 981,58 |

43° | 980,44 | 57° | 981,67 |

44° | 980,53 | 58° | 981,75 |

45° | 980,62 | 59° | 981,83 |

46° | 980,71 | 60° | 981,91 |

47° | 980,80 | 70° | 982,54 |

48° | 980,89 | 80° | 982,98 |

49° | 980,98 | 90° | 983,221 |

##### If g is to be determined more precisely for a higher location, then 0.00031 * H cm must be subtracted from the above table value, where H means the altitude of the location above sea level in meters.

Example: The Altenburg observatory is located at 50 ° 58 '20 "north latitude and 208 m above sea level. By making an intermediate calculation, you can easily determine the exact value of g for Altenburg. According to the table above, g for 51 ° = 981.16 cm / sec^{2} and for 50 ° = 981.07, i.e. a difference of 0.09 cm / sec for one degree^{2}; this is to be distributed over 1 ° = 60 '= 3600 ". Now one second (1") is the 1/3600 th part of 1 ° or: 0.000278 ° = 1 "; for 1" difference in this case 0, 09 / 0.000278 cm / sec^{2} ; from 50 ° 58 '20 "to 51 °, 1' 40" is missing, i.e. 60 + 40 = 100 ". The value you are looking for is therefore obtained if you add the value for 100" from 981.16 (for 51 °), ie Subtract 0.09 * 100 / 0.000278 = 9 * 0.000278 = 0.002502. However, the value is so small that it is not taken into account for reasons given below.

The second larger value to be deducted from Altenburg because of the altitude is 0.00031 * 208 = 0.06448. We therefore have g = 981.16-0.06 = 981.10 cm / sec^{2}.

The following example shows an application of the above types of calculation to the pre-regulation of an astronomical clock for a different installation location:

In Freiburg i.Br. a pendulum clock is to be regulated for Cartago. Freiburg lies at 48 ° N at 298 m above sea level. Cartago lies at 10 ° N at an altitude of 1417 m above sea level. For what time difference should the clock be set?

Freiburg i.Br. | Cartago | |||

48 ° N | g = 980.89 | 10 ° N | g = 978.25 | |

- 0,00031 * 298 | - 0,09 | - 0,00031 * 1417 | - 0, 14 | |

G_{1} = | 980,80 | G_{2} = | 977,81 |

delta t = 44 * (g

_{1}- g

_{2}) = 44 * (980,80 - 977,81) = 44 * 2,99

**= 131.56 sec.**

##### The clock must therefore be in Freiburg i.Br. Proceed 1.31 seconds or 2 minutes 11½ seconds daily.

This is only an approximate calculation, which is usually completely sufficient, because the exact value of g has to be determined through experiments, as it still depends on the composition and nature of the earth's crust at the observation site - ore deposits, caves, etc. - and the latitudes do not Have circular shape, but are ellipses.

The calculation is simpler when it comes to locations that are both between the 40th and 50th degree of latitude, since the change in g between these limits can be assumed to be uniform. For 1 ° it is 0.086 cm / sec^{2} ; the corresponding change in length of the seconds pendulum 0.087 mm. If a pendulum-seconds clock is brought closer to the pole in the specified range, the clock advances by 3.8 seconds on a day (3.8 seconds / d, seconds per day).

If the clock were set up 1 m higher than before, this would result in a lag of 0.011 sec / d.

Task: A pendulum-seconds clock built in Glashütte should go right there when brought to Berlin. How must this be fine-tuned when Glashütte 50 ° 51 '10 "N at H = 330 m above sea level and Berlin 52 ° 30' 17" at H = 37 m above sea level. lie?

Resolution: Glashütte = 50 ° 51 '10 "- Berlin = 52 ° 30' 17" = difference = 1 ° 39 '07 "= 1.652 °Glashütte = 330 m above sea level - Berlin = 37 meters above sea level = Difference = 293 m.

corresponds to: 1.652 * 3.8 = 6.28 sec (width) + 3.22 sec (height)

**= 9.5 sec / d**

##### The clock goes ahead in Berlin by 9.5 sec / d, so it has to be adjusted in Glashütte for a daily retardation of 9.5 sec.

The content of this section can be summarized briefly for a precision seconds pendulum clock in the following sentences:

- The greater the geographical latitude, i.e. the closer to the poles, the longer the pendulum. Taken one degree further north, the same clock runs 3.8 seconds a day
**in front**. - The higher above sea level, i.e. further from the center of the earth, the shorter the pendulum. The clock goes 1 meter higher for 0.011 seconds per day
**to**. - The higher the mean barometer reading, the shorter the pendulum. 1 millimeter more mercury column the clock runs on average 0.015 seconds per day
**to**.

##### Close observation and experiments have led to the fact that there are clocks today that seldom exceed a daily rate difference of a hundredth of a second.

Finally, it should be noted that the art of watchmaking occupies an exceptional position with regard to measurements; for in length, area, weight and the like measurements, which can be carried out as often as desired by the most varied of people with the most varied of aids and also at the most varied of times, the error remains unchanged, ie does not increase; but if the pendulum beat of a clock were only wrong by 1/10000 of a second, one would have a time difference of more than 8 seconds in one day, i.e. an unbearable error, because it is added 86400 times, even immeasurable!

"The technical calculation of the watchmaker", 2nd issue, by H. Romershausen,

Publishing house Richard Markewitz, Mühlhausen / Thuringia (year unknown)

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