A massless spring hangs from the ceiling with a small object attached to its low

Question

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 23 cm below yi. (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is 19 cm below the initial position? (c) An object of mass 280 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 sping?

assless spri gs from the ceiling with a object attached to it ower end. The object itially held at rest position y, su h that the sp at its rest gth. The object is then relea d from y d oscillates up a ith its lowast positi being 23 cm below yi (a) What is the trequency of the oscillation? (b) What is the speed of the object when it is 19 am below the initial position? (c) An object ot mass 280 gis attached to the tirst object, atter which the system oscillates with h the original frequency. What is the mass of the first object? (d) How far below y i tha new equilibrium (rest) position with both objects attached to the sping? (a) Number Lo unit (b) Numb 59 un (c) Num Unit (d) Numb 1.46

Explanation / Answer

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 = 23cm, and A = 11.5cm
F = kx
m·9.8m/s² = .115m·k
k/m = 9.8/0.115s² = 85.22/s²
sqrt(k/m) = 9.23/s
f = 9.23/2s = 1.46/s=1.46HZ (a)

b) max Ep = ½k·(0.115m)² = 6.6125e-3m²·k
at y = 19cm, Ep = ½k(0.095m)² = 4.5125e-3m²·k
E = 2.1e-3m²·k
This must have become kinetic energy, Ek = ½mv²
Set them equal, and solve for v²:
v² = 1.4e-3m²·2·k/m
From above, we have k/m = 85.22 /s²
v = sqrt(2.1e-3m² · 2 · 85.22/s²) = 0..598 m/s

c) sqrt(k/(m + 0.1kg) )= ½sqrt(k/m)
square both sides
k/(m+0.28kg) = k/(4m)
m + 0.28kg = 4m
3m = 0.28kg
m = 0.0933kg

d) we've quadrupled the force on the spring; therefore we've quadrupled the displacement.
4 x115m = .46m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.