Aditi Chauhan
Last Activity: 9 Years ago
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.