DATA:amplitude:(in m/s)bob'strajectoryfrictionless pivotmassless rodequilibriumpositionmassive bobThe pendulum bob is first held at rest atangle = 0o and then released. The bob swingsdown, and then swings back and forth between 8= -80 and 8o. For the data given below,calculate the speed of the bob when 0=0.L = 1.4 m;"Simple gravity pendulum" model assumes nofriction or air resistance.m = arbitrary; 00=39 degrees;OA: 0.386 OB: 0.560 OC: 0.812 OD: 1.177 OE: 1.706 OF: 2.474 OG: 3.587 OH: 5.202

Name: DATA:amplitude:(in m/s)bob'strajectoryfrictionless pivotmassless rode
Uploaded: 2024-03-20T06:58:04.000Z
Channel: Bartleby
Description: DATA:amplitude:(in m/s)bob'strajectoryfrictionless pivotmassless rodequilibriumpositionmassive bobThe pendulum bob is first held at rest atangle = 0o and then released. The bob swingsdown, and then swings back and forth between 8= -80 and 8o. For th

Given data: Length (L) = 1.4 m Angle (θ) = 39° Required: The speed of the bob at the bottom We know that Acceleration due to gravity (g) = 9.81 m/s2 The decrease in height Δh = L - L.cos(θ) According to conservation of energy The total energy of the system is remains constant. Gain in kinetic energy = loss in potential energy ∆KE = ∆PE12mv2 = mg∆hv = 2g∆h= 2g.L1-cosθ= 2×(9.81 m/s2)1.4 m1-cos39°= 2.474 m/s The speed of the bob at the bottom is 2.574 m/s