0.460 kg mass suspended from a spring oscillates with a period of 1.50 s. How much mass must be added to change the period to 1.95 s?
T=2π•sqrt(m/k)
T₁²=4π²m₁/k
T₂²=4π²m₂/k
T₁²/T₂²=m₁/m₂
m₂=m₁T₂²/T₁²
Δm= m₂-m₁ =
=m₁(T₂²/T₁²-1)=
=0.46[(1.95/1.5)² -1] =0.317 kg
To solve this problem, we can use the equation for the period of a mass-spring system:
T = 2π√(m/k)
where T is the period, m is the mass, and k is the spring constant.
We are given the initial period T₁ as 1.50 s, and the final period T₂ as 1.95 s. Let's denote the initial mass as m₁, and the mass to be added as Δm. The final mass would then be m₂ = m₁ + Δm.
We can rewrite the equation as:
T₁ = 2π√(m₁/k)
T₂ = 2π√(m₂/k)
Since the spring constant k is constant, we can compare the two equations:
T₂/T₁ = √(m₂/m₁)
Now we can substitute the known values into the equation:
1.95 / 1.50 = √((m₁ + Δm)/m₁)
To isolate Δm, we square both sides of the equation:
(1.95 / 1.50)² = (m₁ + Δm)/m₁
Simplifying the equation:
(1.95 / 1.50)² * m₁ = m₁ + Δm
(1.95 / 1.50)² * m₁ - m₁ = Δm
Now we can substitute the given mass m₁ = 0.460 kg into the equation:
Δm = (1.95 / 1.50)² * 0.460 kg - 0.460 kg
Calculating the result:
Δm = (1.3)² * 0.460 kg - 0.460 kg
Δm = 1.69 * 0.460 kg - 0.460 kg
Δm = 0.776 kg - 0.460 kg
Δm = 0.316 kg
Therefore, to change the period from 1.50 s to 1.95 s, a mass of 0.316 kg must be added to the system.