My Math Forum A light elastic spring hanging vertically with a mass attached to its lower end.

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 November 15th, 2016, 10:49 AM #1 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 A light elastic spring hanging vertically with a mass attached to its lower end. A light elastic spring has a natural length $a$ and modulus of elasticity $λ$. The energy stored in the spring when it is stretched is $$\frac{λx^2}{2a}$$ where $x$ is the extension. A light elastic spring of natural length $0.2m$ and modulus of elasticity $50N$ hangs vertically with one end attached to a fixed point and with a particle of mass $2Kg$ attached to the lower end. (i) Calculate the extension of the spring when the particle is in equilibrium. (ii) The particle is pulled down below its equilibrium position until the total extension of the spring is $0.2m$ and it is then released from rest in this position. Calculate the speed of the particle when it passes the equilibrium position, and find the maximum compression of the spring in the resulting motion. Answers so far: (i) In equilibrium, $b$ is the static extension and $T$ is the tension in the spring (Hooke's Law): $$T = \frac{λb}{a}$$ Resolving vertically: $$T=mg$$ so $$b=\frac{mga}{λ} = \frac{(2)(10)(0.2)}{50}=0.08m$$ (ii) The total extension of the spring is $(b+x) = 0.2m$, and the spring is released from rest in this position. In the extended position: $$T=\frac{λ(b+x)}{a} = \frac{(50)(0.2)}{0.2}= 50N$$ and using: $$v^2 = ω^2(a^2 - x^2)$$ we have $$v= √ (√125 (0.2^2 - 0.08^2)) = 0.6 m/s$$ where $$ω=√(\frac{λ}{ma})$$ and $x=0.08m$ For the maximum compression of the spring the answer is quoted as $3N$, but should this be in $m$ not $N$? Last edited by Statistics132; November 15th, 2016 at 10:57 AM.
 November 15th, 2016, 11:35 PM #2 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 How was the $3N$ for the maximum compression of the spring was worked out?
 November 18th, 2016, 01:41 PM #3 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 Ignore the part above which says: "and using: $$v^2=ω^2(a^2−x^2)$$ we have $$v=√(√125(0.2^2−0.08^2))=0.6m/s$$ where $$ω=√(\frac{λ}{ma})"$$ as it is wrong. It should say: and using: $$v^2=ω^2(a^2−x^2)$$ with $ω=5 /s$, $a=0.12m$ and $x=0 m$, as we want the velocity as the mass passes the equilibrium position, which is at $x=0m$, so we have: $$v^2 = 5^2(0.12^2 - 0^2) = 0.36$$ and $$v=√(0.36) = 0.6m/s$$ The $ω$ comes from the differential equation of the system, which is: $$\frac{d^2 x}{dt^2}=-25x$$ Last edited by Statistics132; November 18th, 2016 at 02:30 PM.
 November 19th, 2016, 03:10 PM #4 Member   Joined: Jun 2015 From: Warwick Posts: 37 Thanks: 1 The value of the velocity given in my book is wrong. It should be $1.3 m/s$. Also the value for $ω$ is $√125$, so the differential equation for the system should be $$\frac{d^2 x}{dt^2} = -125x$$ $$v^2 = (√125)^2.(0.12^2 - 0^2) = 1.8$$ and $$v= √1.8 = 1.3 m/s$$ Last edited by Statistics132; November 19th, 2016 at 03:14 PM.

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