A 2. 14-kg object hangs from a spring. A 325-g body hung below the object stretches the spring 1.80 cm farther. The 325-g body is removed and the object is set into oscillation. Find the period of the motion.

Step 1: Determine the Spring Constant (k)
The additional 325-g mass stretches the spring by 1.80 cm. Using Hooke’s law:

F = kx

where
F = force due to the additional mass = mg
m = 325 g = 0.325 kg
g = 9.81 m/s²
x = 1.80 cm = 0.018 m

Now, calculate F:

F = (0.325 kg)(9.81 m/s²)
F = 3.18825 N

Solving for k:

k = F / x
k = 3.18825 N / 0.018 m
k = 177.13 N/m

Step 2: Find the Period of Oscillation
The period of a simple mass-spring system is given by the formula:

T = 2π √(m / k)

where
m = 2.14 kg (since the 325-g mass is removed)
k = 177.13 N/m

Substituting values:

T = 2π √(2.14 kg / 177.13 N/m)
T = 2π √(0.01209 s²)
T = 2π (0.1100 s)
T ≈ 0.691 s

Final Answer:
The period of oscillation of the 2.14-kg object is 0.691 seconds.