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?
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.
To solve this problem, we can use the formula for the period of a mass-spring system:
T = 2π√(m/k)
Where:
T = Period
m = Mass
k = Spring constant
Let's start with the given information:
Initial period (T1) = 1.50s
Initial mass (m1) = 0.500kg
New period (T2) = 2.00s
First, we can rearrange the formula to solve for the spring constant (k):
k = (4π²m) / T²
Next, we can calculate the initial spring constant (k1) using the given values:
k1 = (4π² * 0.500kg) / (1.50s)²
k1 ≈ 4.19 N/m (rounded to two decimal places)
Now, we can rearrange the formula to solve for the new mass (m2):
m2 = k2 * T2² / (4π²)
To find k2, we use the formula for the period (T2) with the new mass (m1 + Δm):
T2 = 2π√((m1 + Δm) / k2)
Now we can substitute the values and solve for m2:
2.00s = 2π√((0.500kg + Δm) / k2)
Squaring both sides:
4.00s² = 4π²((0.500kg + Δm) / k2)
Dividing both sides by 4π²:
1.00s² / (4π²) = (0.500kg + Δm) / k2
Rearranging to solve for k2:
k2 = (0.500kg + Δm) / (1.00s² / (4π²))
k2 = (0.500kg + Δm) / (1.00s² * (1/4π²))
k2 = (0.500kg + Δm) / ((1/4) * (1/π²))
k2 = (0.500kg + Δm) / (4/π²)
k2 = π²(0.500kg + Δm) / 4
Substituting k2 back into the equation and solving for m2:
1.00s² / (4π²) = (0.500kg + Δm) / (π²(0.500kg + Δm) / 4)
Cross-multiplying:
4π² = (π²(0.500kg + Δm) / 4) * (1.00s² / 1)
Simplifying:
4π² = (π²(0.500kg + Δm)) / 4
Multiplying both sides by 4:
16π² = π²(0.500kg + Δm)
Dividing both sides by π²:
16 = 0.500kg + Δm
Subtracting 0.500kg from both sides:
Δm = 16 - 0.500kg
So, the mass that must be added to the object to change the period to 2.00s is approximately 15.50 kg (rounded to two decimal places).