- An 83kg student is hanging from a bungee cord that is stretched 5.0m past its unstretched length and then released. - The spring constant of the cord is 270N/m. - The question asks where the student is located and what their velocity is 2 seconds later. - To solve this, the forces and known values are identified. The point of equilibrium is calculated to be 3.0m below the original endpoint. - The student's position and velocity 2 seconds later is calculated to be 1.8m above the equilibrium point (1.2m below the original endpoint) with a velocity of -1.6m/s, meaning they have passed

1. PHYS 101 LD2 – Learning Object 001
Vertical Oscillations Question
Question:
An 83kg studenthangingfroma bungee cordwhois pulleddownwarduntil the cordisstretched 5.0m
past the unstretchedlengthandthenreleased.The springconstantis270N/m. Where isthe studentand
whatis hisvelocity2slater?
Draw out all forces,knowns,andunknownsbefore continuing.
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5.0m
83kg

2. PHYS 101 LD2 – Learning Object 001
Vertical Oscillations Explanation
Question:
An 83kg studenthangingfroma bungee cordwhois pulleddownwarduntil the cordisstretched5.0m
past the unstretchedlengthandthenreleased.The springconstantis270N/m. Where isthe studentand
whatis hisvelocity2slater?
Mstudent = 83kg
K = 270N/m
Cord stretch= 5.0m
t = 2.0s
Althoughthe bungee cordisstretched5.0mwhenthe studentisreleased,thisisnotthe amplitudeof
the oscillation.Oscillationsoccuraroundthe equilibriumposition,sofirstwe needtofindthe pointof
equilibrium,where the studenthangsmotionless.The equation below givesthe cordstretchat
equilibrium:
Whichis equal to3.0m (ΔL = 2.0m).
Therefore acord beingstretched5.0mbringsthe student2.0m below the equilibriumpoint.Fromthis
we can conclude (A = 2.0m),or that the studentoscillateswith amplitude A =2.0m abouta point3.0m
belowthe bungee cord’soriginal endpoint.
The student’spositionasafunctionof time,asmeasuredfromequilibrium, is:
y(t) = (2.5m)cos(ωt+ φo)
where ω = √klm= 1.80 rad/s. The initial condition:
yo = Acosφo = -A
requiresthe phase constanttobe φo = π rad. At t = 2.0s the student’spositionandvelocityare:
y = (2m)cos((1.8rad/s)(2s)(+π rad) = 1.8m (position)
vy = -ωA sin(ωt+ φo) = =-1.6m/s (velocity)
Therefore,the studentis1.8m above the equilibriumposition,or1.2m below the original endof the
cord. Because hisvelocityisnegative,he hastherefore passedthroughthe highestpointandisheading
back down.
Good luck to everyone studying!