Question

A 0.500kg mass suspended from a spring oscillates with a period if 1.50s. How much mass must be added to the object to change the period to 2.00s?

Answer 1

We can use the formula for the period of an oscillating spring:

T = 2π √(m/k)

where T is the period, m is the mass, and k is the spring constant.

Solving for k, we get:

k = (2π/T)^2 * m

Since we are adding mass to the object, the new mass will be m + Δm, where Δm is the added mass. The new period will be T' = 2.00 s. We can set up two equations using the original period and the new period:

T = 1.50 s --> m + 0 = m

T' = 2.00 s --> m + Δm = ?

Using the original period equation, we can solve for the spring constant:

k = (2π/1.50 s)^2 * 0.500 kg
k = 13.20 N/m

Using the new period equation, we can solve for the added mass:

2.00 s = 2π √((m + Δm)/13.20)
(2.00 s/2π)^2 * 13.20 = m + Δm
1.38 kg = m + Δm
Δm = 1.38 kg - 0.500 kg
Δm = 0.88 kg

Therefore, we need to add 0.88 kg of mass to the object to change the period from 1.50 s to 2.00 s.