m₁g=kx
k= m₁g/x
ω=sqrt(k/m₂)=… (rad/s)
f=ω/2π=…(Hz)
A 2.0kg block hang from a vertical spring causes it to stretch by 200cm. If the 2.0kg block is replaced by a 0.50kg mass and the spring is stretched and released, what are the frequency and period of the oscillations?
Please help and provide solutions and explanations. Thank you in advance! Well appreciated and God bless!
3 answers
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The frequency and period of the oscillations can be determined using the following equations:
1. For the initial situation with a 2.0kg block hanging from the spring:
From Hooke's Law, we know that F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
The weight of the block is given by m₁g, where m₁ = 2.0kg and g = 9.8m/s².
Setting F = m₁g and x = 200cm = 2.0m, we can solve for the spring constant:
k = F/x = m₁g/x = (2.0kg)(9.8m/s²)/(2.0m) = 19.6 N/m.
2. For the situation with a 0.50kg mass replacing the 2.0kg block:
We can find the new equilibrium position by equating the weight of the mass to the force exerted by the spring:
m₂g = kx₀,
where m₂ = 0.50kg, g = 9.8m/s², x₀ is the new equilibrium position.
Solving for x₀:
x₀ = m₂g/k = (0.50kg)(9.8m/s²)/(19.6 N/m) = 0.25m.
Since the spring was initially stretched by 2.0m, the extension of the spring when the 0.50kg mass replaces the 2.0kg block is:
Δx = x - x₀ = 2.0m - 0.25m = 1.75m.
3. The frequency of the oscillations can be found using the formula:
ω = sqrt(k/m₂),
where m₂ = 0.50kg and k = 19.6 N/m.
ω = sqrt(19.6 N/m / 0.50kg) = sqrt(39.2 N/kg) ≈ 6.26 rad/s.
4. Finally, the period of the oscillations is given by:
T = 2π/ω = 2π / 6.26 rad/s ≈ 1.00 s.
So, the frequency of the oscillations is approximately 6.26 Hz and the period is approximately 1.00 s.
1. For the initial situation with a 2.0kg block hanging from the spring:
From Hooke's Law, we know that F = kx, where F is the force exerted by the spring, k is the spring constant, and x is the displacement of the spring.
The weight of the block is given by m₁g, where m₁ = 2.0kg and g = 9.8m/s².
Setting F = m₁g and x = 200cm = 2.0m, we can solve for the spring constant:
k = F/x = m₁g/x = (2.0kg)(9.8m/s²)/(2.0m) = 19.6 N/m.
2. For the situation with a 0.50kg mass replacing the 2.0kg block:
We can find the new equilibrium position by equating the weight of the mass to the force exerted by the spring:
m₂g = kx₀,
where m₂ = 0.50kg, g = 9.8m/s², x₀ is the new equilibrium position.
Solving for x₀:
x₀ = m₂g/k = (0.50kg)(9.8m/s²)/(19.6 N/m) = 0.25m.
Since the spring was initially stretched by 2.0m, the extension of the spring when the 0.50kg mass replaces the 2.0kg block is:
Δx = x - x₀ = 2.0m - 0.25m = 1.75m.
3. The frequency of the oscillations can be found using the formula:
ω = sqrt(k/m₂),
where m₂ = 0.50kg and k = 19.6 N/m.
ω = sqrt(19.6 N/m / 0.50kg) = sqrt(39.2 N/kg) ≈ 6.26 rad/s.
4. Finally, the period of the oscillations is given by:
T = 2π/ω = 2π / 6.26 rad/s ≈ 1.00 s.
So, the frequency of the oscillations is approximately 6.26 Hz and the period is approximately 1.00 s.
2 answers