標題:
Applied Maths - Mechanics 42
An endless cord consists of two portions, of length 2a and 2b repectively, knotted together, their masses per unit length being m1 , m2 . It is placed in stable equilibrium over a small smooth peg and then slightly displaced. Show that the time of a complete oscillation is 2pi { (m1 a + m2 b)/[(m1 - m2)g] }... 顯示更多 An endless cord consists of two portions, of length 2a and 2b repectively, knotted together, their masses per unit length being m1 , m2 . It is placed in stable equilibrium over a small smooth peg and then slightly displaced. Show that the time of a complete oscillation is 2pi { (m1 a + m2 b)/[(m1 - m2)g] } ^0.5 .
最佳解答:
WLOG, we may suppose m1 > m2. Then when the cord is slightly displaced for a small length, say x, then: One side will gain a mass of (m1 - m2)x and the other side will lose a mass of (m1 - m2)x. Hence the mass difference between two sides of the cord is 2(m1 - m2)x, giving a net force = 2(m1 - m2)gx. Total mass of the cord = 2(m1a + m2b). Finally, magnitude of acceleration of the cord is: 2(m1 - m2)gx/[2(m1a + m2b)] = (m1 - m2)gx/(m1a + m2b) So: d2x/dt2 = (m1 - m2)gx/(m1a + m2b) implying that it is an s.h.m. Comparing with d2x/dt2 = ω2x: ω2 = (m1 - m2)g/(m1a + m2b) ω = √[(m1 - m2)g/(m1a + m2b)] Then the period is given by: 2π/ω = 2π√[(m1a + m2b)/(m1 - m2)g]
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Applied Maths - Mechanics 42
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發問:An endless cord consists of two portions, of length 2a and 2b repectively, knotted together, their masses per unit length being m1 , m2 . It is placed in stable equilibrium over a small smooth peg and then slightly displaced. Show that the time of a complete oscillation is 2pi { (m1 a + m2 b)/[(m1 - m2)g] }... 顯示更多 An endless cord consists of two portions, of length 2a and 2b repectively, knotted together, their masses per unit length being m1 , m2 . It is placed in stable equilibrium over a small smooth peg and then slightly displaced. Show that the time of a complete oscillation is 2pi { (m1 a + m2 b)/[(m1 - m2)g] } ^0.5 .
最佳解答:
WLOG, we may suppose m1 > m2. Then when the cord is slightly displaced for a small length, say x, then: One side will gain a mass of (m1 - m2)x and the other side will lose a mass of (m1 - m2)x. Hence the mass difference between two sides of the cord is 2(m1 - m2)x, giving a net force = 2(m1 - m2)gx. Total mass of the cord = 2(m1a + m2b). Finally, magnitude of acceleration of the cord is: 2(m1 - m2)gx/[2(m1a + m2b)] = (m1 - m2)gx/(m1a + m2b) So: d2x/dt2 = (m1 - m2)gx/(m1a + m2b) implying that it is an s.h.m. Comparing with d2x/dt2 = ω2x: ω2 = (m1 - m2)g/(m1a + m2b) ω = √[(m1 - m2)g/(m1a + m2b)] Then the period is given by: 2π/ω = 2π√[(m1a + m2b)/(m1 - m2)g]
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