
Applied Math Applied Math Forum 
 LinkBack  Thread Tools  Display Modes 
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. 

Tags 
attached, elastic, end, hanging, light, lower, mass, spring, vertically 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
A particle of mass m attached to a light inextensible string of the length a.  Statistics132  Applied Math  3  November 12th, 2016 01:46 PM 
Initial value & massspring equation problems please help!  heartilly89  Differential Equations  1  April 14th, 2014 09:59 PM 
Stability of a standart massspring system  ronaldinho55  Calculus  0  September 11th, 2013 06:39 AM 
A man of mass 80 kg dives vertically downwards into a swimmi  rsoy  Physics  6  June 23rd, 2012 11:44 AM 
Springmass system  matsorz  Calculus  1  September 5th, 2011 09:12 AM 