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Unformatted text preview: 1.11) Waves On a String String has mass m , length l and tension . The speed of wave has dimensions [ v ] = L T while the quantitites listed above have dimensions [ m ] = M [ l ] = L [ ] = ML/T 2 So to match dimensions [ m l ] = [ v ] we must have L + = 1 M + = T- 2 =- 1 which has a solution = 1 / 2 , =- 1 / 2 , = 1 / 2 and we can conclude v = C radicalbigg l m As we might expect, the speed depends on the ratio of mass to length- a string twice as long of the same material would have twice the mass and, if under the same tension, waves would propagate at the same speed. 1.16) Projectile with Drag Given height as a function of time y ( t ) = 1 parenleftBig v sin + g parenrightBig ( 1- e- t )- gt In the limit of small , the quantity t is nearly zero, and we can perform a Taylor expansion of the exponential function. e- t 1- t + 2 t 2 2!- ... which gives us y ( t ) 1 parenleftBig v sin + g parenrightBig parenleftbigg t- 2 t 2 2! parenrightbigg- gt Multiplying through... y ( t ) v t sin - v t 2 2 sin gt - gt 2 2- gt 1 and cancelling terms reveals y ( t ) v t sin - v t 2 2 sin - gt 2 2 Given that is small, we can assume the second term is negligible compared to the other two, leaving y ( t ) v t sin - gt 2 2 which is the solution in the absence of any drag force. 2.20) Block Under an Overhang a65 a65 a65 a65 a65 a65 a65 a65 a65 a65 a65 a8 a8 M a8 a8 a65 a65 a65 a65 a45 Mg a0 a0 a0 a0 a0 a0 a0 a0 a0 a0 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a64 a72 a72 a72 a72 a72 a72 a89 Mg a1 a1 a1 a1 a1 a1 a11 Mg a54 N Using the plane of the incline as the x-axis, and with the normal force along the y-axis, we get force components F x Mg cos + Mg sin = F f F y Mg sin = N + Mg cos Solving for the unknown forces...
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