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 10 cm below yi. (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is 8.1 cm below the initial position? (c) An object of mass 310 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?

Answer 1

a] Frequency of oscillation will be:

also, at the lowest position, mg = kA

Where A = amplitude = 0.1 m

=> k/m = g/A = 9.8/0.1 = 98

therefore,

b] At x = 8.1 cm, the kinetic energy is:

K = total energy - elastic potential energy

=> (1/2)mv² = (1/2)kA² - (1/2)kx²

=>

=>

c]

f ' = f/2

and

so if the frequency halved while the spring's constant remained the same, the new mass must be four times the initial mass.

(m+0.31) = 4m

=> m = 0.10333 kg = 103.33 g.

this is the mass of the first object.

d]

k/(m+0.31) = k/4m

but k/m = 98 [from above]

therefore, 98/4 = g/A'

=> A' = 0.4 m = 40 cm.

so, the new equilibrium is 40 cm below yi.