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A massless spring hangs from the ceiling with a small object attached to its lower end.The object is initially held at rest in a position yi such that the spring is at its rest length. The object is then released from yi and oscillates up and down, with its lowest position being 10 cm below yi.(a) What is the frequency of the oscillation? (b) What is the speed of the object when it is 8.0 cm below the initial position? (c) An object of mass 300 g is attached to the first object,after which the system oscillates with half the original frequency.What is the mass of the first object? (d) How far below yi is the new equilibrium (rest) position with both objects attached to the spring?

Afaq Masih , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 
We have to assume that "at its rest length" means "at its UNLOADED rest length"; otherwise, it would not need to be held. Then twice the amplitude = 10cm, and A = 5cm 
F = kx 
m·9.8m/s² = 0.05m·k 
k/m = 9.8/0.05s² = 196 /s² 
sqrt(k/m) = 14/s 
f = 14/2πs = 2.22/s ← (a) 
b) max Ep = ½k·(0.05m)² = 1.25e-3m²·k 
at y = 8cm, Ep = ½k(0.025m)² = 3.1e-4m²·k 
?E = 0.94 e-3m²·k 
This must have become kinetic energy, Ek = ½mv² 
Set them equal, and solve for v²: 
c) sqrt(k/(m + 0.1kg) )= ½sqrt(k/m) 
square both sides 
k/(m+0.3kg) = k/(4m) 
m + 0.3kg = 4m 
3m = 0.3kg 
m = 0.1kg = 100 gm
d) we've quadrupled the force on the spring; therefore we've quadrupled the displacement. 
4 · 5cm = 20cm ←
 
Regards
Arun (askIITians forum expert)

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