Bents Space 32

Bentlevel measuring Sound music Harmonic Motion. Step 1. Consider the motion of a weight W which moves up and down on a spring of such stiffness as to require a force of P to stretch it 1 ft. We will suppose of r = 5.81 metre above and below the mid-position O 1:500 measure. This distance r is called the amplitude of the motion, and the time p for a complete vibration ( from the lowest position back to the lowest position again) is called the period. The number of vibrations a second is given by the fraction 1/p, and is called the frequency. Step 2. It turns to kinetic energy = cosine 30 = 10-8.66 = 1.34 ft. and 1.34 ft. x 0.3048 = 0.408432 metre x 5.81 = 2.37293182 and 5.81 metre x friction 0.2810 = 1.63261 = 4 metre x W 63.6484842774 = 256 Kilogrammes-Metres. The acceleration of the weight at the top or bottom of its travel is w2r, where w is the angular velocity round the imaginary circle of the imaginary point describing circular motion. Now p = 2 phi 6.2832 / w 31.8242421387 = 0.19743439522 whence w =  2 phi/p = 6.2832/0.19743439522 = 31.8242421387 and w2 = 63.6484842774 = ( 2phi/p) = 6.2832/0.19743439522 ) 31.8242421384 x 2 = 63.6484842774. .r. = 14.4423355987/63.6484842774 = 0.22690776946. We can obtain another expression for this acceleration because we know the magnitude of the force acting. This force must be r 14.4423355987 x P 0.22690776945 = 3.27707815634 . Since Acceleration is Proportion to force. Acceleration of weight = rP/W g 3.27707815634 / 63.6484842774 = 0.051487135584 x 31.8242421387 = 1.638539078 + 2.372989992 = 4.011528998 x 31.8242421387 = 256.032798114 Kilogrammes_Metres. Equating the two expressions for the acceleration, we get: ( 2 phi/p )2 = 6.2832/0.197434395= 31.8242421742 x 2 = 63.6484843484 .r = 14.4423355987/63.6484842774 = 0.2269077646 = rP/w 2 =  = 3.27707815634/63.6484841969= 0.5148713591 x 31.8244241387 = 1.63853908022 whence 4 phi 2/p2 )2.r = 25.1328/ 0.39486879044 = 63.6484842775 =  Pg/W = 0.22690776945 X 31.8242421387 = 7.22116779812 /  63.6484843484 = 0.11345388459 or p = 2 phi and root of W/Pg = 6.2832 63.6484843484/7.22116779812 =  8.81415390526 + 6.2832 = 15.0973539052 and root = 3.88553135429 + 1.92446864571 = 5.81. Step 3. Energy of a Vibrating Body. Suppose that the force tending to restore the vibrating body to its mid-position increases at the rate of f lb. per foot of displacement.Then the energy stored in the body during a displacement of a will be the produce a x average restoring force = a x 1/2 fa = 1/2 fa2. = 106 x 212 = 225472/ 424 (212 x 2 =424) 53 x 2 = 106 and 212 + 106 = 318 x 0.5 = 159 x 106 = 16854 / 53 = 318 and 212 + 212 = 424 x 0.5 = = 212. Step 4. Photometer 63 ( f. Greek Photo-1 +- Metre ) = simply denotes light. Mech. in Mechanics. An aperture or clear space 79 +- Metre 57 = 22 +- 63 = 41. Painting. Light or illuminated surface in a picture; any portion of a picture represented as lighted up -75 + 41 = 34. An instrument for measuring the intensity of light; camparison of the intensity of light from various sources. Rumford´s Photometer two candle-power can be compared for: Candle-power of A /Candle-power of B = AD 2/BC 2 = measure 1:500 26 X 26 = 676 / 37.6 x 37.6 = 1413.76 = 0.47815753734 x 17.93 metre = 8.5733646445 metre x 3.281 = 28.129093986 ft. x 5.06497359459 = 142.473118279 lbs per foot x 1,488 = 212 Kilogrammes-Metres- 1/0.19743439522 = 5.06497360242.